IMAGE EVALUATION TEST TARGET (MT-3) 4o / A % 7a 1.0 I.I ■ 50 ™^= ■ Si 1^ ,2 It m IL25 i u M 1.8 1.6 "Z .^^ / 'c» A. ,-V'' ^ Hiotographic Sciences Corporation 23 WEST MAIN STREET WEBSTER, N.Y. 14580 (7)6) 87^-4503 ^> *'; .1 THE LIBRARY UNIVERSITY OF WESTERN ONTARIO THE J. D. BARNETT TEXT-BOOK COLLECTION / y ^ 'iSmm^ »»»*' m^^i »,»•#••' - ,.v. -.i -•""•:;:> ■■ -,-,.„.«-.-•»-•--»■ 1'ikm •mamma— •'—!". ,*••.•>•"■»•!•:••;;;? St" t I ,"..ui!.— — --„n•''-?• ^ University of Western Ontario LIBRARY LONDON - CANADA Class \^^^ \ \ ^f?s:i:5Si=5.:^ ..;:,-jli«!»»»f"I ,. i.ii.S-.'""; ■■■■■I* ■•""-"TTr iif<'«(i /or Second Class Certificates and Intermediate Ejcaminutions, ELEMENTS OF GEOMETRY -«& CONTAINING BOOKS I. and II. y WITH ..'■^- h^^^-^^Q:. EXEBCISES AND NOTES. \ BY J. HAMBLIN SMITH, M.A!, of Goimille and Cania CoVecio, mid laie Lecturer at St. Jfettr'a CoUifje, CanibriaQt. ' '"■'- ■,,•'■'■■1"/ "'■■^.l^'i" -.,1/'/ :^"A' ' ,'-% ■ ft' WITH i:;v// 4;'v# SELECTION OF EXAMINATION PAPERS, BY THOS. KIRKLAND, MA., SCIENCE MASTER, NORMAL SCHOOL. CANADIAN COPTBIGHT EDITION. ronoNTO: W. J. GAGE & CO. 1882. ESntered according to Act of the Parliament of Canada, in the year one thousand eight hundred and seventy-seven, by Adam Millek & Co., in the of^ce of the Minister of Agriculture. I9)(i t8 n the n, »>y ter of PEEFACE. To preserve Euclid's order, to supply omissions, to remove defects, to give brief notes of explanation and simpler methods of proof in cases of acknowledged difficulty — such are the main objects of this Edition of the Elements. The work is based on the Greek text, as it is given in the Editions of August and Peyrard. To the suggestions of the late Professor De Morgan, published in the Companion to the British Almanack for 1849, I have paid constant deference. A limited use of symbolic representation, wherein the symbols stand for words and not for operations, is generally regarded as desirable, and I have been assured, by the highest authorities on this point, that the symbols employed in this book are admissible in the Examinations at Oxford and Cambridge.^ I have generally followed Euclid's method of proof, but not to the exclusion of other methods recom- ^ I regard this point as completely set ;led in Camhrirlge by the following notices prefixed to the jiapers on Euclid set in the Senate-House Examinations : I. In the Previous Examination : In ansivei's to these questions any intelligible symbols and abbre- viations may he used. II. In the Mathematical Tripos : In answers to the questions on Euclid the symbol — mmst not be lised. The only abbreviation admitted /or the square on AB is "sq. on AB," and for the rectangle contained by AB and CD, *'rect. AB, CD." vm PREFACE. monded l)y tlicir sinijiluiity, such as tlio demonstration!? l)y which I ]n'oposo to replace (at least for a first read- ing) the difficuh Theorems 5 and 7 in the First Book. I have also attouiptcd to render many of the proofs, as for instance Propositions 2, 1 3, and 35 in Book L, and Proposition 13 in Book II., less confusing to the learner. In Propositions 4, .5, G, 7, and 8 of the Second Book I have ventured to make an important change in Euclid's mode of exposition, by omitting the diagonals from the dia^jrrams and tlie gnomons from the text. In the Third Bouk I have deviated with even greater boldness fi'oin the precipe line of Euclid's method. For it is in treating of tlie properties of the circle that tlie importance of certain matters, to which reference is made in tlie Notes of the present volume, is fully hrouglit out. 1 allude especially to the appli- cation of Superposition as a test of equality, to the conception of an Anglo as a magnitude capable of unlimited increase, and to the development of the methods connected with Loci and Symmetry. The Exercises have been selected with considerable care, chiefly f]'om the Senate House Examination Papers. They are intended to be progressive and easy, so that a learner may from the first be induced to work out something fur himself I desire to express my thanks to the friends who liave improved tlds work by their suggestions, and to beg for further help of the same kind. J. ha:\iblin smith Cambridge, 1873. \ ELEMENTS OF GEOMETRY. INTRODUCTORY REMARKS. When a block of stono is hewn from the rock, \re call it a Solid Bochj. The stone-cutter slinpcs it, and brings it into that vhicn we call regularity of form; and then it becomea a Solid Figure. Now suppose the figure to be snch that t'uo block has six flat sides, each the exact counter- part of thp others ; so that, to one -who stands facin'T a corner of the block, the three sides which are visible present the appearance re- presented in this di;i;:^ram. Each side of the figure is called a Surface ; and when smoothed and polished, it is called a Plane Surface. The sharp and well-defined edges, in which each pair of sides meets, are called Lines. The place, at which any three of the ed^es meet, is called a Point. A Magnitude is anythinr^ which is made up of parts in any way like itself. Thus, a line is a mafinitude ; because we mi; regard it as made up of parts which are themselves lines. The properties Lencrth, Breadth (or Width), and Thickness (or Depth or Height) of a body are called its Dimensions. "We make the following distinction between Solids, Surfaces, Lines, and Points : A Solid has three dimensions, Length, Breadth, Thickness. A Surface has two dimensions, Length, Breadth. A Line has one dimension, Length. A point has no dimensions. s. E. BOOK I. DEFINITIONS. I. A Point is that whicli has no parts. This is equivalent to saying that a Point has no magnitude, since we define it as tliat which cannot be divided into smaller parts. II. A Line is length without breadth. We cannot conceive a visible line without hroadth ; but we can reason about lines as if they had no breadth, and this is what Euclid requires us to do. in. The Extremities of finite Lines are points. A point marks position, as for instance, the place where a line begins or ends, or meets or crosses another line. IV". A Straight Line is one which lies in the same direction from point to point throughout its length. V. A Surface is that which has length and breadth only. VI. The Extremities of a Surface are lines. VII. A Plane Surface is one in which, if any two points be taken, the straight line between them lies wholly in that surface. « Thus the ends of an uncut cedar-pencil are plane surfaces ; but the rest of the surface of the pencil is not a plane surface, since two points may be taken in it such that the straight line joining them will not lie on the surface of the pencil. In our introductory remarks we gave examples of a Surface, a Line, and a Point, as we know them through the evidence of the senses. I » tilde, laller but this ere a etion loints that aces ; [■face, t line rface, ience Book I.] DEFINITIOi . .V. The Siufiices, Lines, and Points of Geometry niny bo rej^arded as mental pieLiirt's of the siufuces, lines, and points which wo know from experience. It is, however, to be observed that Geometry requires us to conceive the possibility of the existenco ot a Surface apart from a Solid body, of a Line apart from a Surface. of a Point apart from a Line. VIII. When two straight lines meet one another, the inclina- tion of the lines to one another is called an Anglk. When tivo straight lines have one point common to both, they are said to Jorm an angle (or angles) at that point. The point is called the ve^rkx. of tho angle (or angles), and the lines are called the a/7/ts of the angle (or angles). O ^ ^ JET Thus, if the lines OA, OB are terminated at the same point O, they form an angle, which is called. y Euclid's deiinition, and the important results that follow an extension of the definition. IX. When a straight line (as AB) meetinj^ another straight line (as CD) makes the adjacent angles {ABC and ABD) equal to one another, each of the angles is called a Eight Angle ; and each line is s:ud to be a Per- pendicular to the other. ^' B D X. An Obtuse Axgle is one which is greater than a right aniile. XI. An Acute Angle is one which is less than a right an tile. XII. A Figure is that which is enclosed by one or more boundaries. Book I.] DLf/Xn/ONS. Xnr. A Ciurrj; is a pliiiic tij^uio cmtiiiiit'd by onu lino, which is called tho (Jircumfkukn'ce, and is such, tluit ull struij/ht lines drawn to tho cireiunforencc from a. certain point (called tho CENxaE) wiihin tho fi^auc arc equal to ono another. XIV. Any straight lino drawn from the centre of a circle to the circnmferenco is called a flAnius. XV. A Diameter of a circle is a atraij^dit lino drawn through the centre; and tenuinatcd both ways by the circumference. Thus, in the diarrram, is the centre of the circle ABCD^ OA, OB, OC, OD are l?adii of the circle, and the stniijilit line ADD is a Diuuieter. Hence the radius of u ciicle is half tho diameter. XVI. A Semicircle is the figure contained by a diamctci and the part of the circumference cut off by the diameter. XVII. IvECTiLiNEAU fij^ures are those which are contained by straight lines. The rEiiiMETEii (or Periphery) of a rectilinear figure is the sum of its sides. XVIII. A Triangle is a plane figure contained by three straijijht lines. XIX. A Ql'adrilateral is a p;;ine figure contained by- four strai^dit lines. XX. A Polygon is a plane figure contained by more than four straight lines. When a polygon h;is nil ils sides equal and all its angles equal it is called a regular polygon. EU CUD'S ELEMENTS. [Book 1. XXI. An Equilateral Triangle is one which has all its sides equal. XXII. An Isosceles Triangle is one which has two sides equal. The third side is often called the 6ase of the triangle. The term hase, is applied to any one of the sides of a triangle to distinguish it from the other two, especially when they have been previously mentioned. XXIII. A EiGHT-ANOLED Triangle is one in which one of the angles is a right angle. The side suhtendiiig, that is, ivhich is opposite the right anale, is called the Hypotenuse. XXIV. An Obtuse-angled Triangle is one in which one of the angles is obtuse. It will be shewn hereafter that a triangle fan have only one of its angles either equal to, or greater than, a right angle. XXV. An Acute-angled Triangle is one in which ALL the angles are acute. XXVI. Parallel Straight Lines are such as, being in the same plane, never meet when continually produced in both directions. Euclid proceeds to put forward Six Postulates, or Requests, that he may be allowed to make certain assumptions on the construction of figures and the properties of geometrical mag- nitudes. Sook I. 3 of a f when b anatle, e only angle. quests, on the il niag- BooK i.] rOSTULA TES. Postulates Let it be granted — I. That a straight line may be drawn from any one point to any other point. II. That a terminated straight line may be produced to any length in a straight line. III. That a circle may be described from any centre at any distance from that centre. IV. That all right angles are equal to one another. V. That two straight lines cannot enclose a space. VI. That if a straight line meet two other straight lines, 80 as to make the two interior angles on the same side of it, taken together, less than two right angles, these straight lines being continually produced shall at length meet upon that side, on which are the angles, which are together less than two right angles. The word rendered " Postulates " is in the original ahfifiara, "requests." In the first three Postulates Euclid states the use, under certain restrictions, which he desires to make of certain in- struments for the construction of lines and circles. In Post. I. and ii. he asks for the use of the straight ruler, wherewith to draw straight lines. The restriction is, that the ruler is not supposed to be marked with divisions so as to measure lines. In Post. III. he asks for the use of a pair of compasses, wherewith to describe a circle, wliose centre is at one extremity of a given line, and whose circumference passes through the other extremity of that line. The restriction is, that the compasses are not supposed to be capable of conveying distances. Post. IV. and v. refer to simple geometrical facts, which Euclid desires to take for granted. Post. VI. may, as we shall shew hereafter, be deduced from a more simple Postulate. The student must defer the consideration of this Postulate, till he has reached the 17th Proposition of Book I, Euclid next enumerates, a^j statements of fact, nine Axioms mi EUCLID'S ELEMENTS. [Book L tlji I m \ ; I I; , or, as he calls thein, Common Notions, applicable (with the exception of the eiirhth) to all kinds of magnitudes, and not necessarily restricted, as are the Postulates, to geometrical magnitudes. Axioms. I. Things which are equal to the same thing are equal to one another. II. If equals be added to equals, the wholes are equal. III. If equals be taken from equals, the remainders are eqnal. IV. If equals and unequals be added together, the wholes are unequal. V. If equals be taken from unequals, or unequals from equals, the remainders are unequal. VI. Things which are double of the same thing, or of equal things, are equal to one another. VII. Things which are halves of the same thing, or of equal things, are equal to one another. VIII. Magnitudes which coincide with one another are equal to one another. IX. The whole is greater than its part. With his Common Notions Euclid takes the groimd of authority, saying in effect, *' To my Postulates I request, to my Common Notions 1 claim, your assent." Euclid develops the science of Geometry in a series of Propositions, some of which are called Theorems and the rest Problems, though Euclid himself makes no such distinction, By the name Thcorcni we understand a truth, capable of denmnstration or proof by deduction from truths previously admitted or proved. By the name FrohJem we nnderstnnd a construction, capable of being effected by the employment of principles of construc- tion previously admitted or proved. A Corollary is a Theorem or Problem easily deduced from, or ett'ected by means of, a Proposition to which it is attached. We shall divide the First Book of the Elements into three aections. The rea^ou fur this division will appear in the course of the work. Book I] SYMBOLS AND ABBREVIATIONS. SYMBOLS AND ABBREVIATIONS USED IN BOOK I. •.' fm because j © jot circle .' therefore qq,q circumference = is (or are) equal to : || parallel L angle CO parallelogram A trianiile .perpendicular equilat equilateral extr exterior intr interior pt point rectil rectilinear reqd required rt right sq square sqq squHies at straight It ia well known that one of the chief difficulties with learners of Euclid is to distinguish lietweon what is assumed, or given, and what has to he proved in some of the Pro- positions. To make tlie distinction clearer we shall put in italics the statements of what lias to Vie done in a Problem, and what has to he proved in a Theorem. TJie last line in the proof of every Proposition states, that what had to be done or proved has been done or proved. The letters q. e. f. at the end of a Problem stand for Quod erat foxiendum. The letters q. e. d. at the end of a Theorem stand for Quod erat demonstrandum. In the marginal references : Post, stands for Postulate. Def. Definition. Ax Axiom. LI Book I. Proposition 1. Hyp. stands for Hypothesis, sn]j)osiiion^ and refers to •omething granted, or assumed to be true. lu L Ul LIUS el em E NTS. [Book 1. Eoo SECTION I. giv i' On the Properties of Triangles, Proposition I. Problem. To describe an equilateral triangle on a givm straight line. ! I I ) i( ' Let A B be the given st. line. It is required to describe an equilat. A on AB. With centre A and distance AB describe BCD. Post. 3 With centre B and distance BA describe ACE. Post. 3 Froni tlie pt. (.*, in wliich tlie s cut one anotlier, draw tlie st. lines CA, CB. Then will ABC be an equilat. A . For •.* A is the centre of © BCD, .'. AC=AB. And •.' B is the centre of © ACE, .-. BC=^AB. Now •.* AC, BC are eiich= AB, .-. AC==BC. Thus AC, AB, BC are all equal, and an equilat. L ALC has been described on A B. Post. 1. Def. 13. Def. 13. Ax. 1. (). E. F. if'^- Boolr I] riwros/Tio.v //. 11 Proposition II. Problem. From a given point to draw a straight line equal to a given straight lint. t- *i Let A be the f^ivcn pt., and BC the given st. line. It is required to draiofratti A a st. line equal to BC. Froni A to B dmw the st. line AB. Post. 1. On AB describe the equilat. a xiBD. I. 1. With centre B and distance BC describe CGTL Post. 3. Produce BB to meet the Qce COTI in G. 7i> With centre J) and distance DC describe GKL. Post. 3. Produce DA to meet the Qce GKL in L. ThenwiUJL=5a For '.' B is the centre of © CG II, .'. BC^BG. And •.• I) is the centre of © GKL, .-. DL=I)G. And parts of these, DA and DB, are equal. .'. remainder ^iy= remainder BG. But BC^BG ; .-. AL=BC. Def. 13. Def. 13. Dcf. 21. Ax. 3. Ax. 1. Thus from pt. A a st. line AL has been drawn = /?r*. Q. E. F. / 13 EUCLID'S ELEMENTS. [Epok I. m r 'I ^ I: 13 ;■ 11 a 4 Proposition III. Puoblkm, From the greater of two given straijld Urns to cut off a part equal to the less. Let AB\)Q the greater of the two given st. lines J.i>, CD. It is required to cut off from AB a part = CD. From A draw the st. line AE=^CD. With centre A iind distance AE describe © EFH, ciittinii; AB in F. Then will ^i^'= CD. For '.' A is the centre of © FFH, .-. AF=AE. But AE:=CD; .-. AF=CD. Thus from AB a part AF has been cut off= CD. 1.2. I- Ax. 1. Q. E. F. Exercises. 1. Shew thill if straiglit lines be drawn from A and B in the diitgrani of Prop. i. to the other point in which the circles intersect, another equilateral triangle will be described on AB. 2. By a constra(;tion similar to that in Prop. in. produce the less of two given straight lines that it may be equal to the greater. 3. Draw a figure for the case in Prop, ii., in which the given point coincides with Ji. 4. By a similar construction lo that in Prop. i. describe on a given straight line an isosceles triangle, whose equal sides shall be each equal to another given straight line. if' 1 1 , Book I.] PRoros/rrox iv. 13 Proposition IV. Theorem. If two triayiglcs have two sides of the one. equal to tico sides of the other, each to each, arid have lilewwe the anrjles contained by those sides equal to one another, they must have- their third sides equal; and the two triangles 7nust be equal, and the other angles must be equal, each to each, viz. those to which the equal sides are opposite. In the Ai:. ARCDEF, let AB=DE, and AC==2.>F, and z BAC= i EDF. Hien must BC=EF and A ABC =--- A DEE, and the other L s, to vhieh the equal sides are opposite, must be equal, that is, L ABC= ^ DEF and l ACB= l DFE. For, if A ABC be applied to a DEF, so that A coincides with D, and AB falls on I)E, then '.• AB=DE, .\ B will coincide with E. And ••• AB coincides with DE, and ^ BAC= l EDF, Hyp. .-. ylC will fall on Di^, Then *.' AC=hF, .'. C will coincide with F. And •.• B will coinciilo with JT, and C with F, .-. BO will coincide with EF ; for if not, let it fall otherwise as EOF : then the two st. lines BC, EF will enclose a space, which is impossible. Post. 5. .■. BC will coincide with and .*. is equal to EF, Ax. 8. and A ABC a DEF, and z ABC i DEF, and I ACB l DFE. Q. E. D. '4 EUCLID'S ELEMEXTS. [Book I. Note 1. On the Method of Superposition. Two geometrical mapjnitudes are said, in accordance with Ax. VIII. to be equal, when they can be so placed that the boundaries of the one coincide with the boundaries of the other. Thus, two straight lines are equal, if they can bo so placed that the points at their extremities coincide : and two angles are equal, if they can be so placed that their vertices coincide in position and their arms in direction : and two triangles are equal, if they can be so placed that their sides coincide in direction and magnitude. In the application of the test of equality by this Method of Sufcrposition, we assume that an angle or a triangle may bo moved from one place, turned over, and put down in another place, without altering the relative positions of its boundaries. We also assume that if one part of a straight line coincide with one part of another straight line, the other parts of the lines also coincide in direction ; or, that stmight lines, which coincide in two points, coincide when produced. The method of Superposition enables us also to compare magnitudes of the same kind that are unequal. For example, suppose ABC and DEF to be two given angles. Suppose the arm BC to be placed on the arm EF, and the vertex B on the vertex Fj. Then, if the arm BA coincide in direction with the arm ED^ the angle ABC is equal to DBF. If BA fall between TjD and EF in the direction EP, ABC is less than DBF. If BA fall in the direction EQ so that ED is between EC^ and EF, ABC is greater than DBF. 1 i [Book I. Cook I.] NOTE IT. t5 ice with that the I of the ) plactd o angles coincide igles arc ncide in kthod of may bu another ndaries. coincide iS of the 3S, which compare example, and the irm ED, ion EP, between NoTK 2. On the Conditions of Eqiutlity of two Triangles. A Triangle is composed of six parts, three sides and three angles. When the six parts of one triangle are equal to the six parts of another triangle, each to each, the Triangles are said to be equal in all respects. There are four cases in which Euclid proves that two tri- angles are equal in all respects ; viz., when the following parts are equal in the two triangles. 1. Two sides and the angle between them. I. 4. 2. Two angles and the side between them. I. 26. 3. The three sides of each. I. 8. 4. Two angles and the side opposite one of them. I. 26. The Propositions, in which these cases are proved, are the most important in our First Section. The first case we have proved in Prop. iv. Availing ourselves of the method of superposition, we can prove Cases 2 and 3 by a process more simple than that em- ployed by Euclid, and with the further advantage of bringing them into closer connexion with Case 1. We shall therefore give three Propositions, which we designate A, B, and C, in the Place of Euclid's Props, v. vi. vii. viii. The displaced Propositions will be found on pp. 108-112, Proposition A corresponds with Euclid I. 6. B I. 26, first part. .-.^ I. 8. n i6 EUCLliys ELEMENTS. [Book L Proposition A. Theorem. 1/ ixoo sides of a triangle be equal, the angles opposite those siilcs must also be equal. IN' i ft •> i FiQ. 2. In the isosceles triangle ABC, let AC=AB. (Fig. 1.) Then must iABC= i ACB Imagine the L ABC to be taken up, turned round, and set down again in a reversed position as in Fig. 2, and designate ihe angular points A\ B', C, Then in l^ ABC, A'C'B', V AB^A'C, and AC=A'B', and z U^C= i C'A'B', .'. L ABC= L A'C'B'. I. 4. But lA'CB'=lACB\ .'. L ABC= L ACB. AX. 1. Q.E.D. Cor, Hence every equilateral triangle is also equiangular. Note. When one side of a triangle is distinguished from the other sides by being called the Base, the angular point op- posite to that side is called the VctUx of the triangle. ^ Book I.J PROPOSi nox n. «7 Proposition B. Tiieoukm. If two triangles have two angles of the one equal to tioo avglcs of the other, each ^> each, and the sides adjacent to the equal angles in each also equal ; then muat the triangUi he equal in all respects. \ "^ In hsABC,Di:F, let z ABC= L DEF, and i ACB= /. DFE, and BC='EF. Then must AB=DE, and AC=DF, and l BAC= l EDF. For if aBEF be applied to ^ ABC, so thafr E coincides with B, and EF tails on BC ; then ••• EF^BC, .'. F will coincide with C ; and •.• - DEF= l ABC, .: ED will fall on BA ; . •. jD will ftill on BA or BA produced. Again, •/ /. I)FE= z A CB, .-. FD will fall on CA ; ,-. D will fall on CA or CA produced. .'. D must coincide with A, the only pt. common to BA and CA. .'. DE will coincide with and .'. is equal to AB, andDP AC, and A EDF i BAC, and aDEF aABC; and .'. the trian(,de3 are equal in all respects. Q. E. D. Cor. Hence, by a process like that in Prop. A, we can prove the following theorem : If two angles of a triangle he equal, the sides which subtc7id them are also equal (''iucl. I. 6.) /: II . II t'-J *' i8 EUClins Il.EMEXTS. fBook I, PKUl'OHlTION ('. TllKOllKM. If tivo triaiKjIcH have the. three sidtH of the one equal to the three ddea of the uthtr, each to each, the triamjlea must be e<^ual in all reapects. Let the three sides of the diS ABC, DEF be equal, each to each, that is, AB=DE, AC=I)F/in\d BC=EF. Then mud the irianijlts he equal in all respects. Iiiiiicrine the L DEF io be turned over and applied to the A ABC, in such a way tliat EF coincides with BC, and the vertex 1) falls on the side of BC opposite to the side on whiuh A falls ; and join AD. Case I. When AD passes throu^^h BC. Then in cABD, v BDr=BA, .: l BAD= l BDA, I. A. And in aACD, v CD=^CA, .: i CAD= l CD A, I. A. .-. sura of L s BAD, C^i>=3um of z s BDA, CD A, Ax. 2. that is, lBAC= lBDC. Hence we see, referring to the oriorinal triangles, that z BAC=^ L EDF. ,'., by Prop. 4, the triangles are equal > ill respects. Book I.J rRorosirio.\ c. Case If. When tho line joining the. vertioea does not pass throu^i^h BG. Then in lABT>, '.' BD=BA, /. i BAD= l BJjA, I. A. And in aACD, :' (JD=CA, .: l , C, at the base of an isosceles triangle, are bisected by the straight lines BD, CD, meeting in D ; shew that BBC is an isosceles triangle. 4. D, E, F are points taken in the sides BC, CA, AB, of an equilateral triangle, so that BD — CE AF. Shew that the triangle DEF is equilateral. 5. In a given straight line find a point equidistant from two given points ; 1st, on the same side of it ; 2d, on opposite sides of it. G. ABC is a triangle l;avirg the angle ^ZJ(7 acute. In BA, or BA produced, find a point D such that BD—CD. 7. The equal sides AB, AC, of an isosceles triangle ABC are produced to points J?" and G, so that AF=AG. JJG' and CF are joined, and H is the point of their intersection. Prove that BH—Cn, and also that the angle at A is bisected hy ATI. 8. BAC, BDC are isosceles triangles, standing on oppo- site sides of the same base BC. Prove that the straight line from A to I) bisects BC at right angles. 9. In how many directions may the line ^^ be drawn in Prop. III. ? 10. The two sides of a triangle being produced, if the angles on the other side of the base be equal, shew that the triangle is isosceles. 11. ABC, ABD are two triangles on the same base AB and on the same side of it, the vertex of each triangle being outside the other. If AC=AD, shew that BC cannot -£1>.° 12. From C any point in a straight line AB, CD is drawn at right angles to AB, meeting a circle described with centre A and distance AB in D ; and from AD, AE is cut off =.4C: shew that- A FB is n -i-vht anp-lo. Bools I.] PROrOSITJOA' XIII. 25 Proposition XTII. Theorem. The migles which one straight line makes with another upon one side 0/ it are either two right angles, or together equal to two right angles. Fig. 1. Fig. 2. U u B 7) Let AB make with CJD upon one side of it the / s ABC, ABD. Then must these he either two rt. l s, or together equal to two rt. l s. First, if /. ABC= i ABD as in Fig. 1, each of thein is a rt. i . Def. D. Secondly, if / ABC be not= i ABD, as in Fig. 2, from B draw BE ± to CD. I. 11. Then sum of / s ABC, ABD=s\}m of z s EBC, EBA, ABD, and sum of z s EBC, EBD=^mn of l s EBC, EBA, ABD ; .-. sum of z s ABC, ABD^zwm of z s EBC, EBD ; Ax. 1. .'. sum of z s ABC, ABD=^\\xa of a rt. z and a rt. z ; .'. z s ABC, ABD are together=two rt. z s. Q. E. D. Ex. Straight lines drawn connecting the opposite angulnr points of a quadrilateral figure intersect each other in 0. Shew that the angles at are together equal to four rigl^t angles, KoTE (1.) If two angles together make up a right angle, each is called the Complement of the other. Thus, in fig. 2, z ABD is the complement of z ABE. (2.) If two angles together make up two right angles, each is called the Supplement of the other. Thus, in both figures, z ABD is the supplement of z ABC. I 26 E UC LID'S ELEAfLlNlS. [Book I, I Proposition XIV. Theorem. If, at a point in a straif/ht line, firn other straight lines, iipnn th^e opposite sides of it, wahe the arljacent angles together equal to two right avgles, these two straight lines must be in one ayid the sanie straight line. At the pt. B in the st. line AB let the st. lines BC, BD, on opposite sides of AB, ninke z s ABC, ABD together = two rt. angles. Then BD must he in the same st. line icith BG. For if not, let BE be in the same st. line with BC. Then z s ABC, ABE together=two it. i s. I. 13. And z s ABC, ABD together = two rt. z s. Hyp- .'. sum of z s ABC, ABE =smn of z s ABC, ABD. Take away from each of these equals the z ABC ; then z ABE= z ABD, Ax. 3. that is, the less = the greater ; which is impossible, '. BE is not in the same st. line with BC. Similarly it may be shewn that no other line but BD is in the same st. line with BC. * .'. BD i5 in the same st. line with BC. Q. E. D. t Ex. Shew the necessity of the words the opposite sidss in the enunciation. • , 1^ Book I.J }*R0P0S1T10N XV. 27 Proposition XV. Theorem. // two straight lines cut one another, the vertically opposite angles must be equal. Let the st. lines AB, CD cut one another in the pt. E. Then must l AEC= l BED and l AED=^ l BEG. For ••• AE meets CD, .-. sum of z s A EC, AED= two rt. z s. I. 13. And •.' DE meets AB, .'. sum of z s BED, AED =two rt. z s ; I. 13. .-. sum of z s AEC, AED=mm. of z s BED, AED ; .-. z AEC= L BED. Ax. 3. Similarly it may be sbewu that z AED= z BEC. ii. E. D. Corollary I. From this it is manifest, that if two straight lines cut one another, the four angles, which they make at the point of intersection, are together equal to four right angles. Corollary II. All the angles, made by any number of straight lines meeting in one point, are together equal to four right angles. Ex. 1. Shew that the bisectors of AED and BEC are in the same straight line. Ex. 2. Prove that z AED is equal to the angle between two straight lines drawn at right angles from E to AE and EC, if both lie above CD. Ex. 3. If AB, CD bisect each other in E ; shew that the triangles AED, BEC are equal in all respects. til* i\ n\ i I N{ f 28 EUCLIiyS ELEMENTS. [Book L Notes. On UuclicTs defLuition of an Angle. Euclid directs us to regard an angle as the inclination of two straioht lines to each other, which meet, but are 7iot in the same straight line. Thus he does not recognise the existence of a single angle equal in magnitude to two right angles. • The words printed in italics are omitted as needless, in Def. VIII., p. 3, and that definition may be extended with advantage in the following terms : — Def. Let WQE be a fixed straight line, and QP a line which revolves about the fixed point Q, and which at first coincides with QE. IS Then, when QP has reacli^c-d the position represented in the diiigram, we say that it has describtd the angle EQP. When QP has revolved so far as to coincide with QW^ we say that it has de.sciibed an angle cqiial to two right angles. Hence we may obtain an easy proof of Prop. xiu. ; for what- ever the position of PQ may be, the angles which it makes with WE are together equal to two right angles. ' ■•'• Again, in Prop. xv. it is evident that z AED= i EEC, since each has the same supplementary z A EC. We shall shew hereafter, p. 149, how this definition may be extended, so as to embrace angles go-eater than two right angles. • Book 1. 1 I'iwposrnuy x i v. ^ riioPosiTioy XVI. TiiEonKM, If one sitk of a triavgle h/i prorJnced, Vie crterior angle is greater than either ol' tJiC iitUrlor opjioslU anglei. Let, the side BC of a AW ho prnrlnnerl to D. Then must / ACD he greater than either l CAB or l ABC. Bisect AC\u E, nnd join BE. I. 10. Produce BE to F, niulaii-,' EF=BE, and join FC. Then in ls BEA, FEC, :• BE=-FE, and EA ==EC, and /. BEA^ l FEC, I. 15. I EC. lEAB. Now that IS. z A.CD is 'greater than z ECF ; z ACD is o'leater than z i^^li?, z ^CI> 13 nreater than z C^i?. 1.4. Ax. 9. Similarly, if AC be produced to G it may l^e shewn that md z i>C(r is Greater than z = z JCi> ABC. I. 15. z J. CD is L'tenter than z ^Z? o. Q. E. D. Ex. 1. From the same point there cannot he drawn more than two equal straiuht lines t« nieet a given siraiuht line. T.x. 2. If, from any point, a striiitjjht line he drawn 1o ii given Rtraii'ht line making with it an acute and an fhtnse angle, and if, f:'nm the pa-ne point, a perpendicular he drawn to the* given line ; the perpendicular will fall on the side of the ac: e ai.?'e i 1 30 /• U CUD'S L I.EiML XJo. [Uooli 1 Puoi'osrnoN' XVM. Thkoumm. Any two angles 0/ a licauyU ara together less than two right angle$. any Then must any two 0/ its l s be together less than t rt. L 3. Til Produce BC to D. L ACD is tfreiiter than i ABC. Wi) I. le. •. z s ACD, ACB are toj^etht-r (greater tliiin z s ABC, ACB But z &ACD, ACB ti)i;etlier=two it. z s. I. i:i .*. z fiABC, ACB are together less than two rt. z s. Simihirlv it niav be shewn that ifiABC, BAC ami alsc> tliat z sBAC, ACf^ are touether less than two rt. z s. NoTic 4. 0/?. ^/i(; ^/x//j, Postulate. Q. E. D. We learr from Prop. xvii. that if two straij^ht lia.-.? BM aiul ON, which meet in A, aro mut by another straij^ht. line DE iu the poiiit.> 0, JP, the angles MOP ana JVPO are together less than two right anijles. The Sixth Postulate asserts that if a line DE ineetinjT tv/o other li/ieh /M/, C'..V m .Ices MOP, NPO, th« two iut«rior Book i J /'A'on)6//'Jo.v xi'Ji. 3f nnjiles on the bhiiio sIHp of it, to«ret.lier less than two riu'it angles, jBM and CVmI'II nicpt if pioilncfd on the same 8ido of DE on which are tiie ungies ^lOi' and Nl*0. PnorosiTiov XVIII. Theorem. If one side of a triavgh he (jiraler than a tcxond, th> angle opposite the Jirst must be greater than that opponte the In lABC^ let side AC he greater than AB. ^ Then rn.i(st l ABC he nr cater th-an L ACB. From AC cut off AD = A JJ, iind join BD. Then '.' AB=^AD, .'. L ADB= L ABB, And •.• CD, a side of tsBBC. is produced to A. .'. L ABB is oi-euter than t ACB ; .".also z ABD is greater than i ACB. Much more is z ABC greater than z ACB. ' I. 3. I. A. I. 16 Q. IS. D. Ex. Shew that if two angles of a trianple he equal, the sides which subtend them are equal also (h'ucl. I. ^). J RUCLID'S r.LHML.'^n. [BocU L / PUOPUSITIU.V XIX TlllSOlii:!!. // on« an^U of a triangU bi greater tJuni a second, the tidn opposite tha fint )WMt bs greater than that oppotite the In A ABC, let z ABC bu greater than z ACB. Then must AC be greater titan AD. For if AC be not ijreater than AB, AC inu3t either sa-.^lii, or be less than AB, Now .46' cannot = /li?, for then I. ▲. ii A EC would =3 z ^l CB, \\ hich is not the case. And AC cannot be less than AB, for then I. 18. I ABC would be less than z ACB, which is not the case : 4' ^C is greater than AB, Q. E. D. Ex. 1. lu an obtusp-angled triangle, the greatest aide is opposite the obtuse angle. Ex. a. BC, the base of an isnscelrs triangle BAG, i"^ pro- duced to any point I) ; shew that AD is greater than AB. Ex. 3. The perpendicular is the shortest straifdit line, which can be drawn from a given point to a given straifiht line ; and of others, that which is ncr.rer ir, the perpendicular is less ihan ore niora fftniot©. toSs L] rsu>ros/no,v xx. 53 '. ROP'SfTION XX. TilKORr-.M. Anij tiJO aMCi of a inanfjU arc together grcaUr than ths third .ride Lfit ABC ha a. A. Theit. any two of iij iidca must be together greater L\a.r> the third side. Prodaco JJA to D, uvAViw^ AD = AC, and julu 1)C. Then V AD=AC, .'. L ACI)=' L AJjC, that is, z BLG. I. x. Now ii BCD is ;;reater thau ^ ACD , .'. L BCD is iilsu f,n cater thnii l EDC ; •. BD is greater than LC. L 13. But BD=BA and AD torrether ; that \A,BD = BA and AC together ; .*. BA and AC to^'cther aie greater than BC. Similarly it mny be shewn that AB and BC together are greater than AC, and i^Caiid CI AB. Q. K. D. Ex. 1. Prove that any threo sides of a quadrilateral figure are together greater than the fourth side. Ex. 2. 6he\v thiit i:r-y fide of a triangle is greater than the diflereviGc between the oth<^«' ^'^vo sides. Ex. 3. Prove that the snni of the distances of any point from the angular points of a qundrilatcral is greater thau half the perimeter of the quadrilateral. Ex. 4. If one side of a triangle be bisected, the sum of the two other sides shall be more tlian double of the line joining th« vertex and the point of bisection. * I :i 34 /■:cvL//ys /:/J::meats. [Book T. Proposition XXI. Theorem. J/, from the ends of the side of a triavgle, there be drawn two straight lines to a j)orpi within the triangle; these tnll he togetJur Ier,0'int in a given straight line, to make an angle equal to a given angle. Li't A he the ^vp-n pt., rC the giren line, DEF the given / . Tt is reqd. to wale at f f. A an angle «= z DEF. In ED, EF take nnv pts. D. F ; '.\v.(\ join DF. In AB, prorlnced if necessary, majce AG=DE. In AC, prorlnced if necessary, make ATI—EF. In HC, produced if necessary, make TIK=FD. With cer.tre A, and distance AG, describe © GLM. With centre JI, and distance UK, describe qLKM. Join AL and TIL. Then •.' LA^-AC, .: LA =DE ; " Ax. 1. and •.• IIL^IIK, .-. IIL^FD. Ax. 1. Then in a s LAII, DLF, '.■ LA^DE, and AJl^EF, and IIL=FD ; .-. iLAn== lLEF. Lc .*. an anslc LAII Las l:ccn ir.ado r,t pt. Ai as vras reqd. Q. K. p. iiOOll ic] PKOFCj/noy p. 37 Note. — We liere cfivs the proof Oi n theorem, necessary to the proof of Prop. aXIV. and applicable to . I Js i if 38 /•:ucLiD s 1;:llme.\ ts. [Book I. Proposition XXIV. Tiieghem. Jf two trinvgks have tvo sides of the one equal to two sides of the other, each to each., hut the angle contained by the two sides of one of them greater than the angle contained by the two sides equal to them of the other ; the base of that which has the (jreater angle mud be greater than the base of the other. Ip^the £^hABC,DEF, let ^B=DE and AC=JJF, and let z BAC he grenter than z EDF. Then must BC be greater than EF. Of the two sides DE, DF let DE be not greater than DF* At pt. I) in St. line EI) make z EDG^ z BAC, I. 23. and make DG==AC or DF, and join EG, GF. Then:- AB=DE, and AC=DG, and z BAC-= z EDG, .-. BC=EG, Again, -.DG^DF, :. .DFG^ lDGF; .-. z EFG is greater than z DGF ; much more then z EFG is greater than l EGF ; .-. EG is greater than EF. BntEG = BC', .•. BC is greater than EF. Q. E. D. *This line was aflded by Rimson to olivioto a defect in Euclid's proof. Without this condition, three distinct cases miist be discussed. With the condition, we can prove that F must lie below EG. For since Z'/'is n6t less than f)E, and DO is drawn equal to OF, DO is not less than DE. Hence by Prop, d, any line drawn iVoni D to meet EG is less than DG, and thord'oi-e DF, bciuir equal to DG, must extend beyond EG. For another method of provinj; the Proposition, see p. 113. 1.4. I. A. I. 1!). t^. Book I.J rROFOSITlOX XX V. 39 . A. . 10. nT\ ♦5'.. ■wfROPofcirrioN XX V^. Tiikoukm. If two triangles have two sides of tke one equal to tivo sides of the other, each to each, but the base of the one greater than the base of the other ; the angle also, contained by the sides of that which has the greater base, must be greater than the angle contained by the sides equal to them of the other. In the A8ABrj,DEF, let AB--=DE and AC=DF, and let BC be greater than EF. Then must l BAC be grcattr than l EDF. For L BAC is greater than, equal to, or less than z EDF. Now I BAC cannot = z EDF, for then, by i. 4, BC would = EJ^ ; which is not the case. And z BAC cannot be less than z EDF, for then, by i. 24, BC would be less than EF ; which is not the case ; .-. z BAC must be greater than z EDF. Q. E. D. Note. — In Pro}), xxvi. Kuclid includes two cases, in which two triangles are equal in all respects ; viz., when the following parts are equal in the two triangles : 1. Two angles and the side lii-t'.vei.'n (.hem. 2. Two angles and the side ojiposite ono cf them. Of these we have already pioved the first case, in Prop. B, second case left, to form ti ^ly 3ubj( of Prop. XXVI., -which we shidl prove by the method of superposition. For Euclid's proof of Pi. .p. xwi , seo pp 114-115. 40 EUCLID'S ELEMENTS. [Boole I. Proposition XXVI, Theorem. If t'xc triangles have tv:o aiigJa cf the one equal to two a'ngles of the other, each to cadi, a. id o.ie side (qnal io one side,, those sides icing opposite io equal ciugUs iit, each ; then 7nu*t thi iricngles be eqauL in, all respects. lit L ABC^ L DEF, nnd j. ACB= i JDFE, ?«nd AB=>DE: Then must BC^EF, and AC=DF, and iBAC^ i EDF. Suppose A DEF to be npplied to a ABC, so that D coincides with A, and JDE falls on AB. Then •.• DE=AB, .-. iiwill coincide with B ; nnd •.• z DEF= l ABC, ,'. EF will full ou BC. Then must F coincide with C: for, if not, let F fall hclLveeri B and C, at the pt. 11. Join AE. Then '.• z AHB^ z DFE, I. 4. .-. iAEB= lACB, the extr. z = the intr. nnd opposite z , which is impossible, .'. F does not fall between B :ind C. Similarly, it niav be shev/n that F docs not fall on BG produced. .*. F coinuiQcs with C, and .*. BC=EF ; .'. AC^DF, and /: BAC=- i EDF, and.', the triangles are equal in all respects. 1.4. Q. E. D. Book I.] MISCEJJ.ANEOUS EXERCISES. 41 Misreilaiwoiis Exrvif^rs nv Prodis. /. tn XXVT, 1. M is tlio inM ilo point of tlio l)n«o BC of an ipn?cples triangle ABC, and N is a pnint in AC. ^i^liew that the diflerence betwppn AB and AN. MB and MN is less than that between 2. ABC is a triancjle, and tlm anjrle at A is bisected by a straight line which meets BC :it D ; shew that BA is greater than BD, and CA preuter than CD. 3. AB, AC are 8traici[ht lines nieeting in A, and J) is a given point. Draw through D a stniioht line cutting off equal parts from A B, A. C. 4. Draw a straifrht line tlironcrh a^fiven pnint, to make equal angles with two given straight lines which meet. 5. A given angle BAC'is bisected ; if CA be produced to and the angle BAG bisected, the two bisecting lines are at right angles. 6. Two straight lines nre drnwn to the base of a trianjjle from the vertex, one bisecting the vertical angle, and the other bisecting the base. Prove tliat the latter is the greater of the two lines. 7. Shew that Prop. xvii. may be proved without pro- ducing a side of the triangle. 8. Shew that Prop, xvitt. may be provoi! by means of the following construction : cut ofll" AJ) — AB, draw AE, bisecting Z BAG and nu-eting BC in iv, and join DM. 9. Shew that Prop. xx. can be proved, without producing one of the sides of the triangle, by bisecting one of the angles. 10. Given two angles of a triangle and the side adjacent to them, construct the triangle. 11. Shew that the perpendicular?, let fall on two sides of a triangle from any point in tlie straight line bisef'ting the angle contained by the two side.-,, are equal. I u pii i i ■ 1 if \: 4a EUCLliys El EMEXTS. .jvjC* Ii We concliidu Section I. with the proof (omitted by Kaclinj of another ca' ABC, DEF be equal in all resi-ects. For if J C be not =-DF, make AG=DF ; and join BG. Then in ^s BAG, EJJF, :■ BA=ED, and AG=DF, and z BAG= . EDF, .\BG^ EF and z A GB = z LFE. I. 4. But BC=EF, and .'. BG=BC; .'. iBCG=^lBGC. La. First, let i ACB and z DFE be both acute, then z AGB is acute, and .'. z BGC is obtuse ; I. 13. .'. z BCG is obtuse, which is contrary to the hypothesis. Next, let I ACB and z DFE be both obtuse, then z AGB is obtuse, and /. z BGC is acute ; I, 13, ,'. z BCG is acute, which i^ contrary to the hypothesis. ^ok l.) PROPOSinON K. 43 l^astly, let one cf the third angles ACB^ DFE be a right liTigle, If i ACB be a rt. z , then /. BOO is also a rt. z ; I. A. .*. Z8 J5C'C= alternate i DCB. I. 29 Then in i^% ABC. BCD, ■.' AB=CD, and BC is common, and / ABC= l DCB, .'. AC=BD. and / ACB= l DBC. I. 4. Then '.* BC, meeting AC and BD, makes the alternate z s ACB, 7)7?Cequal, /. AC h\\ to BD. Q. E. p. i I 5« EUCr.ID'S ELEMENTS, [Book L MuceUanmus Exercisca on, ^Sections I. and II. 1. If two exterior unijlos of a triatii^le be bisected by straii^lit lines which meet in 0; prove that the perpeiidicuhirs from oii the .sides, or the fides produced, of the triuuyle ore equal, 2. Trisect a rifjht C, is equal to half the ddferiMice between the an<,'les at Jt and 0. 6. If the straij^dit line Af> bisect the anudo at A of the trianj^le -i'i/:J<'7, and />'/>/■/ be drawn perpendicidar to AD, and meeting J. C, or AC produced, iu E ; shew that BU ia equal to DE. 7. Divide a right-angled triangle into two isosceles tri» angles. 8. AB, CD are two given straif;ht lines. Through a point ^between them draw a straight line GIUI, such that the in- tercepted portion GJi shall be bisected in E. 9. The vertical angle of a triangle OPQ is a right, acute, or obtuse angle, according as Oli, ilie line bisecting BQ, is equal to, greater or less than the half of BQ. 10. Shew l)y means of F,x. how to draw a perpen- dicular to ii given str-aght line fro;u ita eUreuiity without pro- ducing it. Book I.] EUCl.IUS ELEMENTS. <1 SECTION III. r On the Equality of Rectilinear Figures in respect of Area. m i 14 The amount of space enclosed by a Figure is called the Area of that figure. Euclid calls two figures cytta/ when they enclose the same amount of space. Tliey may be dissimilar in shape, but if the areas contained within the boundaries of the li^nies be the same, then he calls the figures equal. He regards a triangle, for example, as a figure having sides and angles and area, and he proves in this section that two triangles may have equality of area, though the sides and angles of each may be unequal Coincidence of their boundaries is a test of the equality of all geometrical magnitudes, as we explained iu Note 1, page 14. In the case of lines and angles it is the only test : in the case of figures it is a U&t, hut not the ouli/ teat ; as we shall shew in this Section. The sign =, standing between the symbols denoting two figures, must be read is equal in area to. Before we proceed to prove the Propositions included in this Section, we must complete the list of DeHuitions required in Book I., coutiuuing the numbers prefixed to the definitions iu page 6. pi 58 r.ccLiD'5 i:l ements. [Book I. Definitions. XXVII. A Parallelooram is a. four-sided figure whose opposit* sides are parallel. For brevity we often designate a parallelogram by two letters only, which mark opposite anjjiles. Thus we call tho ri;i;ure in the margin the parallelogram AQ. XXVI II. A Rectangle is a par- allelogram, having one of its angles a right angle. Hence by I. 29, all the angles of n rectangle nre right, angles. XX iX. A Rhombus is a par- allelogram, ha,ving its sides equal. / XXX. A Square is a paral- j- lelogram, having its sides equal and one of its angles a right ; angle. , Hence, by I, 29, all the angles oi a square are right angles. XXXI. A Trapezium is a four-sided figure of which two sides only are parallel. XXXII. A Diagonal of n fnur-sided figure i« ih<» straijjhi llnp joining two of tl)(» oppo«:itr> nrsiT-nlsr .points. Book I.] KXEJ^CISES ON DEflNrriOXS 27-33. 59 XXXin. The Altitude of a Parallelogram is the perpen- dicular distance of cue of its sides from tho side opposite, re;^arded as the Luse. The altitude of a triangle is the perpendicular distance of one of its angular points from tlie side opposite, regarded as the base. * Thus if ABCD bo a parallelogram, and AE a perpendicular let fall from A to CD, AE is the altitude of the parallelogram, and also of the triangle ACD. If a perpendicular be let fall from B to DC produced, meet- ing DC in F, EF is the altitude of the parallelogram. EXEItCISES. Prove the following theorems : .1. The diagonals of a square make with each of the sidei un ani'le equal to half a right angle. 2. If two straight lines bisect each other, the lines joining their extremities will form a paiallelogrurn, 3. Straight lines bisecting two adjacent angles of a paral- lelogram intersect at riglit angles. 4. If the straight lines joining two opposite angular points of a jjaralielogram bisect the angles, the parallelogram has all its sides equal. 5. If the opposite angles of a quadrilateral be equal, the quadrilateral is a parallelogram. 6. If two opposite sides of a quadrilateral figure be equal to one another, and the two remaining sides be also equal to ont another, the figure is a parallelogram. 7. If one angle of a rhombus be equal to two- thirds of two right angles, the di:;goiial drawn from that angular point divides the rhombus into two equiluteral triangles. i 11 6o LUtJJD'S LLEA1L.\ 'J'S. [Book I. pROrOSlTION' XXXIV. TllF.OUEM. Tm op^osUe fliiZ''.? and anqlra of a faralUlogram are equal to one anotJw, and thi diagonal bisects it. 1.25 1.29. O J) Let ATiDC bo a O, aiul BC a rlingonnl of the O Then must AB=DC and A 0= DB, aiil I BAC= L CDB, and z ABD=^ l ACD and lABC=lDCB. For •.* A B is |1 to CD, and BC meets Ihcm, .-. L u4J5C= alternate z DCB , and •.• AC is |1 to BD, and U(7 meets them, .*. i. ^CJ5=iilterniito z DBC. Then in as ABC, DCB, ••• z yl/?l^= z jyCA and z ACB^ z DZ?C, and UC is co;nnioa. a side adjiccn!" to the equal z s in each ; .'. AB^DC, nnd AC--=DB, rnd z i}^C« z CDjB, and aABC-^ a DCB. I. b. AbD ••• z .4i;,':- z DCD, and z i)/}C= z ACB, .'. L 3 ^/?C, DBC together = z s /)CZ?, ACB together, tliatis, L ABD=. lACD. Q. E. D. Fa*. 1. Shew that the diagonals of a parallelogram bisect each other. Ex. 2. Shew that the din jonals of a rectangle arc equal. look I. Book I.J pRorosiTioN XXX y. 61 yiual to I. 2S I. 20. Proposition XXXV. Theorem. Parallelograms on the same bass and between the same parallels are equal Let the Os ABCD, EBCF be on the same base BC and between the same [js AF, BC. Then m^ust ZZ7 ABCD=EJ EBCF. Case I. li AD, EF have no point common to both, Then in the as FDC, EAB, V extr. I FDC=intv. i EAB, and intr. z DFC=extr. l AEB, and DC=AB, .'. lFDC^aEAB. Now O ABCD with a Fi)C*=figure ^BCf ; and ZZ7 iJBCi?' with a ^.45= figure ABCF ; .'. O ABCD with a FDC=CJ EBCF with a i;^J5; I. 29. I. 29. 1.34. I. 26. ^ '^ ^'K Case II. If the sides AD, EF overlap one another. the same method of proof applies. \ I 62 EUCLID'S ELEMENTS. [Book 1. i Oasr hi. If the sides opposite to BC be terminated in tlie same point D, 4 Xk JT the same method of proof is applicable, but it is easier to reason thus : Each of the Os is double of a BDC ; .: £J ABCD=CJ DBCF. I. 34. Vi. K. D. Proposition XXXVI. Theorem. Parallelograms on cqval bases, and between the same ■parallels, are equal to one ati other. Then Let the Os ABCD, EFGH be on equal bases BC, FG, and between the siime ||s yill, BG. Then must O A BCD =01 EFGH. Join BE, CH. vBC=FG, audEH=FG; .\BC=EE; and BC is || to EII. .-. EB is il to CH ; .*. EBCII is a parallelogram. Now O EBCH^ CJABCD. Hyp. 1.34. Hyp. 1.33. 35. they are on the same base ^6* and between the same I|s And £J EBCH=nJEFGH, I. 35. they are on the same base iv/f and between the same ||s CJABCD^CJ EFGH. Q. BL B. [Book 1. Iiated in Book I.] PROPOSITION XXXVII. 63 Proposition XXXVII, Theoresl Triangles upon the same base, and between the same parallels, are equal to one another. same Let A 8 ABO, DBC be on the oame base EC and between the same |Is AD, BC. Then must a ABC= a DBC. From B draw BE || to CA to meet DA produced in E. From C draw CF \\ to BD to meet AD produced w F. Then EBCA and FCBD are panillelogrums, and O EBCA = O i^CL^D, I. 35. •/ they are on the same base and between the same |js. Now A ABC is hnlf of CJ EBCA, 1. 34. and A DBC is half of C7 FCBD ; J. 34. .-. i^ABC^lDBC. Ax. 7. Q. E. D. Ex. 1. If P be a point in a side AB of a parallelogram ABCD, and PC, PD be joined, the trianjrles PAD, PBC are together equal to the triangle PDC. Ex. 2. If A, B be points in one, and C, D points In another of two paridlel straight lines, nnd the lines AD, BC intersect in E, then the triani^les AFC, BED are equal. 'li' ■' 111 l,h 64 EUCL/jyS ELEMENTS. [Book L Boo Proposition XXXVIII. Theorem. Triangles upon equal hoses, and between the sam.e pa/ralleh, are equal to one another. of Let AS ABG, DBF be on equal bases, BC, EF, and between the same ||s BF, AD. Then must A ABC^ A DBF. BVom B draw BG II to GA to meet DA produced in O. From F draw FH \\ to ED to meet AD produced in B.. Then CG and EH are parallelograms, and they are equal, •.* they are on equal bases BG, EF, and between the same Us BFy GH. I. 36 Now A ABG is half of O GG, and A DEF is half of O EH ; .'. lABG^ lDEF. Ax. 7. Q. E. D. Ex. 1. Shew that a straight line, drawn from the vertex of a triangle to bisect the base, divides the triangle into two equal parts. Ex.2. In the equal sides AB, J. C of an isosceles triangle ABG -pomta D, E are taken such that BD— AE. Shew that the triangles GBD, A BE are equal. ^y 1'/^ ■[Book L Book I.J PROPOSITION xxxrx. 65 ralleh Proposition XXXIX. Theorem. Equal triangles upon the same base, and upor^. the sa/me aide of ity are between the same parallels. ^y and ual, ' same 1.36 Lx. 7. D. ertex ) two mgle that v Let the equal a s ABO, DBG be on the same base BGy and on the same side of it. Join AD. Then must AD be \\ to BO. FoT if not, through A draw ^ || to BO, so as to meot BD. or BD produced, in 0, and join 00. Then '.' La A BO, OBO are on the same base and between ihe same ||s, .-. A ABO= A OBC. I. 37. But A ABO= A DBG ; Hyp. .-. l0B0=-aDB0, the 1 688= the greater, which is impossible ; .'. ^Oisnot II to jBC. In the same way it may be shewn that no other line passing through A but AD is || to BO ; .-. AD is 11 to BO Q. £. D. Ex. 1. AD is parallel to BO ; AG, BD meet in E ; BO is produced to P so that the triangle PEB is equal to the triangle ABO : shew that PD is parallel to AG. Ex. 2. If of the four triangles into which the diagonals divide a quadrilateral, two opposite ones are equal, the quad- rilateral has two opposite sides parallel. 8. W. 5 sVI I ji:: I 66 EVCI.JD'S ELEiMENTS. [Book I. .1 ROPOSITION XL. TnEonEM. Equal triangles upon eqval haMst, in the same straight line, and towards the na/nie ^arts, arc bdwecn the same parallels. Let the equal a s ABC, DEF he on eqiial hanes BC, EF in the same st. line BF and towards the same parts. Join Jl). Then must AD he || to BF. For if not, through A draw AO \\ to BF, so as to meet ED, or ED produced, in 0, and join OF. Then A ABG= a OEF, v they are on equal bases and between the same Us. L 38. But t,ABG= aDEF; Hyp. .-. L OEF = A DEF, the le8s=the greater, which is impossible. .•*A0 is not II to BF. In the same way it may be shewn that no other line passing through A but AD is || to BF, .-. AD is II to BF. Q. B. D. Ex. 1. The straight lino, joining the points of bisection of two sides of a triangle, is ]juiallel to the base, and is equal to half the base, Ex. 2. The straight lines, joining the middle points of the sides of a triangle, divide it into four equal triangles. look I, Book 1.] FKOFOSJTJOA XLI. 6.7 1 line. \ EF tUD, i and I. 38. Hyp. SSUlg D. )n of ai to the Proposition XLI. Theorem. If a 'paralhlograiii and a tria>igh he vjjon the same base, and between the saiae jjaralicU, the 2^cii'<^il'<^(oFJ5J right angles, have also the sides AB, -<-lC equal to DE, DF, each to each ; shew that the triangles are e(|ual in every respect. Note. This Theorem has been already deduced as a Co- rollary from Prop. E, page 43. Ex. &. Divide a given straight line into two parts, so that the square on one port shall be double of the square on the other. Ex. 6. If from one of the acute angles of a right-angled triangle a line be drawn to the opposite side, the squares on that side and on the line so drawn are together equal to the sum of the squares on the segment adjacent to the right angle and on the hypotenuse. Ex. 7. In any triangle, if a lino be drawn from the vertex at right angles to the base, the difference between the stjuares on the sides is equal to the difference between the squares on tho sejfments of the ba-se. S y 76 EUCLID'S ELEMENTS. [Book I. l.-i PROPoaiTioN XLVIII. Theorem. If the square lUscribed n-jjon one of the sides of a triangle be equal to the squares described iipon the other two sides of it, the angle contained by those sides is a right angle. Let the sq. on BC, a side of A ABC, be equal to the sum of the sqq, on A B, A C. Then must l BA C be a rt. angle. From pt. A draw AD L to AC. I. 11. Make AD=AB, and join DC. Then '.' AD=AB, ,'. sq. on AD = &(\. on AB ; I. 46, Ex. 2. add to each sq. on AC . then sum of sqq. on AD, AC—awxw of sqq. on AB, AC. But '.' / DAC is a rt. angle, .', sq. on D6'=sum of sqq. on AD, AC \ I. 47. and, by hypothesis, sq. on J5C'=sum of sqq. on AB, AC ; .*. sq. on i)C'=sq. on BC ; .: DfS=Ba I. 46, Ex. S. Tlienin AS ABC, ADC, •: AB=AD, and ACu common, and BC^DC, .: ^BAC=^ A DAC; I. a and z DA is a rt. angle, by construction ; .•. z BAC is a rt. anp;le. Q. E. D. BOOK IL I. G INTRODUCTORY REMARKS. TiiK geometrical figure with Avhich we are chiefly concerned in this book is the Rectanglp:. A rectangle is said to be con- tained by any two of its adjacent sides. Thus if ABCD be a rectangle, it is said to be contained lay AB, AD, or by any other pair of adjacent sides. We shall use the abbreviation red. AB, AD to express the words "the rectangle contained by AB, AD." We shall make frequent use of a Theorem (employed, but not demonstrated, by Euclid) which may be thus stated and proved . Proposition A. Tiii:oitEM. If the adjacent sides of one rectangle he equal to the adjacent sides of another rectangle, each to each, the rectangles are eqval in area. Let ABCD, EFGH be two rectangles : and let AB=EF and BC=^Iroduced in F. Then yl^=suni of AU and CK Now ^^=rect. AB, CB, :' BE^-CB, AD^xect. AC, CB, .' CD=CB, CE=8q. on CB. .'. rect. AB, Ci5=sum of rect. AC, CB and sq. en CB. (l K. D, Note. When a straight line is cut in a point, the distances of the point of section from the ends of the line are called the segments of the line. If a line AB be divided in C, AC and CB are called the internal segments of AB. If a line ^C be produced to JB, AB and CB are called the external segrments of AC. I Book 11] PROPOSITION IV 8i Proposition IV. Theorem. // a straiglit line be divided into aymj two parts, the square on the whole line is equal to the squares on the two parts together icith twice the rectangle contained by the parts. A & . B H F i> G s Let the st. line AB be divided into any two parts in C. Then must sq. on AB=sum of sqq. on AC, CB and twice red. AC, CB. On AB describe the sq. ADEB. I. 46. From AD cut 0^ AH = CB. Then HI) = AC. Draw CO \\ to AD, and HK \\ to AB, meeting CG in F. Then \BK=AH, .: BK^CB, Ax. i. .-. BK, KF, FC, CB are all equal ; and KBC is a rt. i ; .-. CK is the sq. on CB. Def. xxx. Also HG = sq. on A C, : • HF and HD each =^AC. Now .4E=sum of HG, CK, AF, FE, AE=?>(\. on AB, and HG=&(i. on AC, Ci:=sq. on CB, AF=rect AC, CB, FE =vect. AC, CB, CF=CB, FG=ACsindFK=CB. .-. sq. on ^£=sum of sqq. on AC, CB and twice rect. AC, CB. Q. E. D. Ex. In a triangle, whose vertical angle is a right angle, a straight line is drawn from the vertex perpendicular to the base. Shew that the rectangle, contained by the segments of the base, is equal to the square on the perpendicular. S.E. « W> ^H 1*1 m !■!] u 8a /: UC LID'S ELEME.\TS. iBook n. ! Proposition V. Theorem. IJ a straight line he divided into two equal parts and also into two unequal parts, the rectangle contained by the uyiequal . parts, together mth the square on the Uric between the points of section, is equal to the square on half the line. C D 2J K M W Let the st. linp AB be divided equally in C nnd unequally inZ). Th^^n must rect. AD, DB together icith sq. on CD—sq. on CB. On CB describe the sq. CEFB. Draw DG \\ to CE, and from it cut offDH=DB. Draw HLK || to AD, and AK \\ to DJI. 1. 4(j. I. :n. I. -M. Then rect. J^^^rect. AL, Also LG=sq. on CD, ■.•BF=AC,ixndBD = CL. '.'LH^CD, and HO^CD. Then rect, AD, DB roirether with sq. on CI> = AH tonrcther with LO =suni of AL and CH and LO = sum of DF and CH and LG ^CF =sq. on CB. Q. s. D. / Book II.] PROPOSITION VI. 83 Proposition Vi. Theokkm. If a straiyht litie he bisected and produced to any point, the rectangle contained by the whole line thus produced and the part of it p>rodnccd, toyether iri.th the sqnare on half the line bisected, is equal to the square on the straiyJit line which, is made up of the half and the part jirodacni. I. 4(j. I. 31. I. 31. ^ IT Q :f M Let the «t. line AB be bisected in C and produced to D. Then must rect AD, DB together with sq. on CB—sq. on CD. On CD describe the sq. CEFD. I. 4^. Draw BG \\ to CJ^, and cut off BH^BD. I. 31 Throufrh H druw KLM \\ to AD I. 31. Through A draw AK\\ to CE, Now •.• BG= CD and BH=BD ; .-.HG^CB', Ax. 3. .-. rect. MG=^\vot. AL. IT- a. Then rect. AD, DB torrethoi' with sq. on CB = suin of ^'lAf and LG :^.sum oi AL and CM and LG =sum of MG and CM and LG ^CF .•=sq. on CD. Q. E. 1). f.\ Ill )'v^o>^.y''f ' [iiook IL NoTK. Wo here give the proof of an iniportitni theorem, which is usiuilly phiced as a corollary to Proposition V. !l ' i l< ! Proposition B. Tiieouem. 'Die differeitce between the squares on any two strai'jht lines IS equal to the rectangle contained by the, sum and difference of those lines. ni^js J5 ,/ jf/T Let AC, CD be two st. lines, of which JO is the ^'renter, and let them be placed so as to form one st. line AD. Produce AD to B, nutking CB=^AC. Then JD=the sum of the lines AC, CD, and Di) = the dillerence of the lines AC, CD. Then must difference between sqq. on AC, CD=rect. AD, DB. On CB describe the sq. CEFB. I. 46. Draw DG \\ to CE, and from it cut otf DH=DB. I. 33 . Draw HLK n to AD, and AK \\ to DH. I. 31. Then rect. i)i^= rect. AL, .'BF=AC, and BD=CL. Also LG = sq. on CD, '. ■ LH= CD, and HG == CD. Th§p d fFerence between sqq. on AC, CD — difference between sqq. on CB. CD -simi of CH and DF =sum of CH and AL =AH =rect. AD, DH =rect. AD, DB. Q. E. D. Ex. Shew that Propositions V. and VI. might be deduced from this Proposition. , . yf^ ,.. 'A X: 'V. sqn to tO'Jf Vr m t( fom n.} t'/?OPO'>r/vo.\ 11 f. 85 Proposition' VII. TiiKunEM. tx> straight line be divided into auij two parts, the squares on the ivhole line and on one of the parts are eqval to tioice the redrtnale contained by the. uhole and that part together icitlt the square on the other part. ^ \i ■ B (Jr L. C .) ■%/ 'V .X IC D I. 4G. Let AB be divided into any two parts in C. Th'^ must aqq. on AB, BC= twice rect. AB, BC togdiier with sq. on AC. On AB describe the sq. ADEB. From ADcwto^ AR^CB. Draw CF \\ to AD and HGK \\ to AB. I. 31. Then HF=s^. on AC, and Civ=sq. on CB. Then sqq. on AB, BC=^nm of AE and CK =sum of AK, HF, GE and CK =^su2n of AK, HF and CE. Now ^ii:=rect. AB, BC, '■' BK==BC • CE=vect. AB, BC, •/ BE=AB ; HF=sq. on ^4C*. .*. sqq. on AB, £C=t twice rect. AB, .BC together withsq. on AC. Q. E. D. Ex. If straight lines be drawn from G to B and from (J to D. shew that BGD is a straisl.t line. fl\ ' ■'■''y'l m m !i 86 E UCL lis S EL EMENTH. [Book II. ^i\ iBi Proposition VI n. Tiieoui:m. If a straight line he divided into any tuo parts, four times the rectangle contained by the tchuk line and one of the jjart/6, together vnth the square on the other jiart, ia eqval to the square on the straight line ■xhlch is made up of the whole and the Jirst part. A... 2£ T Ji M, Q^^ £^..JL JT 11 JV L P Let the st. line .4B be divided into iiny two parts in G. Produce AH to D, so that BD^BC. Then must four times net. AB, BC together with sq. on AC~sq. on AD. On AD describe the sq. AEFD. From AE eiu. off AM and MA' eaeh = CjB. Through C, B draw C/J, BL \\ to AE. Throuoh M, X draw MGKN, XFllO || to AD. Now •/ XE'^AC, and XP~AC, .: Xil=sq. on AC. I. 46. 1.31. 1.31. Also AG=MP=PL=RF, and CK=GR=BN=KO; .'. sum of these eight rectangles --four tinie,s the sum of AG, CK —four times AK esfour times rect. AB, BC. Then four times rect. AB, BC and sq. oi\ AC = sum of the eight rectauoles and XB. =^AEFD =«gq. on AD. II. A. 11. A. Q. E. D. Book II.l pRoros/'i'iox IX. «7 1.31. I. 31. II. A. II. A. ! Proposition IX. Theorem. If a straight line be (f hided into ttvo equal, and aJ-^o into two unequal pard^, the /tquarcs on the t)ro unequal parts are together double of the square on half the line and of thf square on the line betuxm the points of section. Let AB he divided equally in Cand unequally in P. Then must funi of sqq. on AD, DB = iivice sum of sqq. on AC, CD. Draw CE-=^ACut rt. /. ^ io AB, and join F.A, EB. Draw DF at it. z b to AB, iiieetinrf ES in F. Draw FG at rt. z s 1o EC, and join AF Then •.• l ACE is a rt. .: , ,'. sum of z s A EC, EAC=a, rt. /. ; and ••• I AEC= l EAC, .: z.4isT'-=balfart. i. So also z BEC and z EBC are each = half a rt. z . Hence z AEF is a rt. z . I. .3fi. I. A. Also, '.• z GEF i^i half a rt. z , .'. z 'EFG is half a rt. z ; .-. z EFC= L GEF, and .-. EG^GF. So also z BED is half a rt. z , and BD=DF. and z EGF is a rt. z ; I. B. Cor. Now sum of sqq. nn A D, DB =sq. on AD together with sq. on DF =sq. on AF I. 47. — sq. on AE too;ethcr with nq. on EF I. 47. ^^^nq. on ^C, iiJC torrether with sqq. on EG, GF I. 47 = twice sq. on AC to^.^ether with twice sq. on GF — twice sq. on AC together with twice sq. on CD. Q. K. p, IMAGE EVALUATION TEST TARGET (MT-3) // 1.0 I.I 1.25 1^ i^ IIIIIM 1^ II 2.2 H: U& 12.0 1.8 U IIIIII.6 .%. '^W/ #■ >5V# 7 Hiotographic Sciences Corporation 23 WEST MAIN STREET WEBSTER, N.Y. 14580 (716) 87^-4503 #^ V \\ ^J.=sum of sqq. on BC, CD, DA and twice rect. BC, CD. Now sqq. on BD, DA=sq. on AB, and sqq. on CD, i>-4=sq. on CA ; 1.47. 1.47. .'. sq. on ^J5=sum of sqq. on BC, CA and twice rect. BC, CD. .*. sq. on AB is greater than sum of sqq. on J5C, CA by twice rect. BC, CD. Q. E. D. Ex. The squares on the diagonals of a trapezium are together equal to the squares on its two sides, which are not parallel, and twice the rectangle contsiined by the sides, which are parallel. Book n.] PROPOSITION XIII. 91 PuorOSITIOX XIII. TlIEORElI. In cury triangle, the square on the side subtending any of the acute amjles is km than the squares on the sides containing that angle, by tivice the rectangle contained by either of these sides and the straight line intercepted hdween the perpendicular, let fad upon it from the opposite angle, and the acute angle. Fio. 2. ij c Let ABC be any a , having the i ABC acute. From A draw AD A. to BC or BC produced. Then must sq. on AC be less than the sum of sqq. on AB, BC, by twice red. BC, BD. For iu Fig. 1 BC is divided into two parts in D, and in Fig. 2B.D is divided into two parts iu C; .'.in both eases sum of sqq. on BC, BD=s,\xm of twice rect. BC, BD and sq. on CD. II. 7. Add to each the sq. on DA, then sum of sqq. on BC, BD, DJ.=sum of twice rect. BC, BD and sqq. on CD, DA ; .'. sum of sqq. on BC, J,J5==sum of twice rect. BC, BD and sq. on ^1 C ; I. 47. .'. sq. on J. (7 is less than sum of sqq. on AB, BC by twice rect. BC, BD. The case, iu which the perpendicular AD coincides with J.C, needs no proof. Q. E. D. Ex. Prove that the sum of the squares on any two sides of a triangle is equal to twice the sum of the squares on half the base and on the line joining the vertical angle with the middle point of the base. I Mi I 1)11 J* !li 92 EUCLID'S ELEMENl'S. [Book II. Proposition XIV. Problem. To describe a square that shall be eqvxxl to a given rectilijiear figure. Let A be the given rectil. fio^re. It is reqd. to descrioe a square that shall = A. Describe the rectangular O BODE = A. I. 45. Then if BE=ED tlie O BCDE is a square, and what was reqd. is done. But if BE be not = ED, produce BE to F, so that EF=ED. Bisect BF in G ; and with centre G and distance GB, describe the semicircle BHF. Produce DE to if and join GH. Then, •.• BF is divided equally in 6r and unequally in E, .'. rect. BE, EF together with sq. on GE =sq. on GF II. 5. =sq. on GH =suni of sqq. on EH, GE. I. 47. Take from each the square on GE. Then rect. BE, EF^sq. on EH. But rect. BE, EF==BD, '.' EF=ED ; .'. sci. on EH =BD', .'. sq. on EH=Tecti\. figure A. Q* £. F. V Book II. Book n.] MISCELLANEOUS EXERCISES. 93 >•{ ctilinear 1.45. ance OB, in JE J 11. 5. 1.47. Q. E. P. Miscellaneous Exercises on Booh II. 1. In a triangle, whose vertical angle is a rigni angle, a straight, line is drawn from the vertex perpendicular to the base ; shew that the square on either of the sides adjacent to the right angle is equal to the rectangle contained by the base and the segment of it adjacent to that side. 2. The squares on the diagonals of a parallelogram are to- gether equal to the squares on the four sides. 3. If ABGD be any rectangle, and any point either within or without the rectangle, shew that the sum of the squares on OJ., OG is equal to the sum of the squares ou OB, OD. 4. If either diagonal of a parallelogram be equal to one of the sides about the opposite angle of the figure, the square on it shall be less than the square on the other diameter, by twice the square on the other side about that opposite angle. 6. Produce a given straight line AB to 0, so that the rect- angle, contained by the sum and difference oiAB and ACy may De equal to a given square. 6. Shew that the sum of the squares on the diagonals of any quadrilateral is less than the sum of the squares on the four sides, by four times the square on the line joining the middle points of the diagonals. 7. If the square on tiie perpendicular from the vertex of a triangle is equal to the rectangle, contained by the segments of the base, the vertical angle is a right angle. 8. If two straight lines be given, shew how to produce one of them so that the rectangle contained by it and the produced part may be equal to the square on the other. 9. If a straight line be divided into three parts, the square on the whole line is equal to the sum of the squares on the parts together with twice the rectanule contained by each two of the parts. w EUCLID'S ELEMENTS. [Book U. 10. In any quixlrilateral the squares on the diagonals are together equal to twice tlie sum of the squares on the straight lines j(»ining the middle points of opposite sides. 11. If straight lines be drawn from each angle of a triangle to bisect the opposite sides, four times the sum of the squares on these lines is equnl to three times the sum of the squares on the sides of the triangle. 12. CD is drawn pei-pendicular to AB, a side of the triangle ABG^ in which AC—AB. Shew that the square on CD is eijual to the square on BD together with twice the rectangle AD, DB. 13. The hypotenuse AB of a right-angled triangle ABC'xa trisected in the points 1>, E\ prove that if VI), C'ii\ be joined, the sum of the i^quiires on the sides of the triangle VDB is equid to two-thirds of the square on AB. 14. The square on the hypotenuse of an isosceles right angled triangle is equal to four times the square on the perpendicular from the right angle on the hypotenuse. 15. Divide a given strnight line into two parts, so that the rectangle contained by them shall be equal to the square described upon a straight line, which is less than half the line divided. I« '4^ ook It. Books I & II.] OJV THE MEASUREMENT O/- AREAS. 9c uls are itraight iiiangle squares lares on triangle I CD ia Bctangle ABC'iA L^ joined, (WE is t angled ■ndicular go that 10 square the Una Note 6. — On fhe Meaf^nremenf of AreoM. To measure a iVTa<»nitudp, we fix upon some magnitude of the same kind to serve as a standard or unit ; and then any magnitude of that kind is measured by the number of times it contains this unit, and this number is called the Mkasure of the quantity. Suppose, for instance, we wish to nieasure a straij^ht line AB. We tiike another straight line EF for our standard, B C D E F and then we say if ^ B contain EF three times, the measure of ^jB is 3, if four 4, if X X. Next suppose we wish to measure two straight lines AB, CD by the same standard EF. If AB contain EF m times and CD n times, where m and n stand for numbers, whole or fractional, we say that AB and CD are commensv-aMe. But it may happen that we j.iay be able to find a standard line EF, such that it is contained an exact number of times in AB ; and yet there is no number, whole or fractional, which will express the number of times EF is contained in CD. In such a case, where no unit-line can be found, such that it is contained an exact number of times in each of two lines AB, CD, these two lines are called incommensurable. In the processes of Geometry we constantly meet with incommensurable magnitudes. Thus the side and diagonal of a square are incommensurables ; and so are the diameter and circumference of a circle. ^1 i 96 EUCLID'^ ELEMENTS. [Books I. & II. Next, suppose two lines vl/>', A(' to be itt tij,^lit jmf,'les to each other ami to bo commensuiablo, so that AB contains four times a certain unit of linear measurement, wliich irf contained by AQ three times. A il c .. 2) Divide AB, AC into four and three equal parts respectively, and draw lines through the points of division parallel to AC, AB respectively ; then the rectangle ACDB is divided into a number of equal squares, each constructed on a line equal to the unit of linear measurement. . If one of these squares be taken as the unit of area, the measure of the ar(?a of the rectangle ACDB will be the number of these squares. Now this number will evidently be the .same as that obtained by multiplying the measure of AB by the measure of AC; that is, the measure oi AB being 4 and the measure of AC 3, the measure of ACDB is 4 X 3 or 12. (Algebra, Art. 38.) And generally, if the measures of two adjacent sides of a rectangle, supposed to be commensurable, be a *nd fc, then the measure of the rectangle will be oh. (Algebra, Art. 39.) If all lines were commensurable, then, whatever might be the length of two adjacent sides of a rectangle, we might select the unit of length, so that the measures of the two sides should be whole numbers ; and then we might apply the processes of Algebra to establish many Propositions in Geometry by simpler methods than those adopted by Euclid. Take, for example, the theorem in Book ii. Prop, iv. If all lines were commensurable we might proceed thus . — Let the measure of ^C be ^r, of 05 ... y. Then the measure of J B is .r+?/. Now (x + yr = •n'^ + ?/^ + S.r'/. which proves the theorem. I. & II. Books!. £i IT) OX rill': MEASrREMEXT or AREA.''. 97 ijjllo.s to ns four [itained actively, L to AC, i into a equal to rea, tlio number FiUf. iniisnmcli as all linos are nut coniiiicnsiirjible, vo have in Cicometry to treat of mmjniimlcs. and not of mra^ure.'i : that is, when we use the syuibol A to represent a line (as in I. 22), A stands for the lino itself and not, as in Algebra, for the number of units of length containod by the line. The method, adopted by Euclid in Book II. to explain the relations between the rectnufflos containod by certain lines, is inoie exnct than any methofl fonndod upon Algebraical prin- ciples can be ; because his method applies not merely to tho oiise in which the sides of a rectangle are commensurable, but also to the case in whii^h thoy are incommensurable. The student is now in a position to tinderstand the practical iippiication of the theory of Equivalence of Areas, of which the foundation is the 35th Proposition of Book I. We shall give a few examples of the use made of this theory in Men- sir rati on. obtained of AC ; >f ACS, 38.) des of a then the n.) ht be the elect the bould be cesses of y simpler bus : — Area of a Parallelogram. The area of a parallelogram ABCD is equal to the area of the rectanijle ABEF on the same base AB and between the same parallels AB, FC' Now JiPj is the altitude of the para!leIogram ABCD if ABhe taken as the base. Hence area of O J.i?CD = rect. AB, BE. If then the measure of the base be denoted by 6, and altitude fe, the measure of the area of the O will be denoted by bh That is, when the base and altitude are commensurable, measure of area = measure of base into measure of altitude. s. E. 98 EUCUaii LLliMENTS. [Boolc3 I. & II. Arm, of a Triangle. If from one of the angular points A of ii triHii^He ABi\ a perpendicular ADhQ drawn to BC, Fij^. 1, or to BC produced, Fig. 2, FlO 2. and if, in both cases, a parallelograui ABCE be completed of which ABj BC hvg adjacent sides, area of a ^J50=half of area of O ABCE. Now if the measure of BC be h, and AD... h, measure of aren of £U A BCE is hh ; .'. measure of area of A ABC is ~ . Area of a Rhombus. Let ABCD be the jrivpn rhombus. Draw the diagonals ytTand BT), cutting one another in 0. It is easy to prove tiiat A C and BD bisect each other at right angles. Then if the measure of AC be x, and BD ... y, measure of area of rhombus = twice measure of a ACD, = twice -:- 4 XII Uoolta I. & II j A A'/. J ol- A J i^APiuliUM. 99 Arm of a Trapezium.. Let ABCD be the given trupeziuiu, having the sides ABf CD piiriiUel. Djuw AE ut right ungles to ClJ. •in 0. Produce DO to F, making CF-=AB. Join AF, ciittinj,' EC in 0. Tlienin La AOB, COF, .• ^ i^.40= z C'i'O, and / .405= z J^OC, and AB^CF) .\ lCOF= aAOB. 1.26. Hence trapezium ABCI)^ aAJjF. Now suppose the measures of AB, CD, AE to be w, 7o,p respectively ; .'. measure oiDF=m + n, '.' CF=AB, Then measure of area of trapezium = A (measure of DF X measure of AE) =^{vi+n)X2). That is, the measure of the area of a trapezium is found by multiplying half the measure of the sum of the parallel sides by the measure of the perpendicular distance between the parallel sides. 1: too Jii 'Cl.JJiS LLKMKN'J'S. [Books I. & II. Area of an Irrcyiilar Fohjcjon. There are throe met lutds of fiiuling the urea of an irreguhir polygoTi, Avliich we shiill here briefly notice. I. The polygon ■luaij be divided into triangles, and the tirei> of each of these triani^les be found separately. Thus the area of the irregular polygon ABODE is equal to the sum of the aretis of the triiingles ABE, EBB, DBG. II. The polyg^ni ■1110,11 he converted into a single triangle of equid area. If ABCDE be a ])ent;igon, we can convert it ijito an equivalent qiiadi'ihiteral by the l(jllovving process : Join BD and draw CF parallel to BD, meeting ED pro- duced in F, and JL)in BF. Then will quadrilateral ^i^J^i!;= pentagon ABCDE. For i\ BDF= /\BCD, on snnie base BD and between same parallels. If, then, from the pentagon we remove a BCD, and add lBDF to the reni;iinder, we obtain a quadrilateral ABFE eqiivalent to the ]tf'ntagoii A T'CDE. Books I. & n.l AAUiA OF.LV iKREGUl^lR /O/.YGO.V. lor The quadrilatei'.il may then, by a siu'iiliir process, be con- verter' into an cqnivalent trianj^^le, and tlius a poly<,^on of any number of sides may be gTadiially converted into an equiva- lent triangle. The area of this triann;le may then be found. III. The third method is cliiefly employed in practice by Surveyors Let ABCDEFG he an irretrular polvo-on. Draw AE, the lonjje.st diagonal, and drop perpendiculars on AE from the other anj^ular points of tlie polygon. The polygon is thus div'ded into figurest Avliich are either right-angled triangles, rectangles, or trapeziums ; and the areas of each of these figures may he readily calculated. I02 EUCLID' :S LLEMEN'l 'S. [Books I. &< 11. Note 7. On. Frojections. The projection of a point B, on a straight line of unlimited length AEy is the point M at the foot of the perpendicular dropped from B on AE. The projection of a straiyht line EC, on a straight line of unlimited length AE, is MN, — the part of AE intercepted between perpendiculars dniwn from B and 0. When two lines, as AB and AE, form an angle, the pro- jection of AB on AE is AM. a ir -Sf We might employ the term projection with advantage to shorten and make clearer the enunciations of Props, xii. and XIII. of Book 11. Thus the enunciation of Prop. xii. might be : — " In oblique-angled triangles, the square on the side sub- tending the obtuse angle is greater than the squares on the sides containing that angle, by twice the i-ectangle contained by one of these sides and the projection of the other on it." The enunciation of Prop. xiii. might be altered in a similar manner. I. & il. [ilimited ndicular t line of ercepted the pro- mtage to XII. and side sub- B5 on the jontained n it." a similar Books I. & II. OjV loci. 103 Note 8. On Loci. Suppose we have to determine the position of a point, which is equidistant from the extremities of a given straight ime^BC. ^ ' There is an infinite number of points satisfying this con dition, for the vertex of any isosceles triangle, described on BQ as its base, is equidistant from B and C. Let ABC be onz of the isosceles triangles described on BC. If BC be bisected in D, M'N., a perpendicular to BC drawn through Z>, will pass through A. It is easy to shew that any point in 3fiV, or MN produced ill either direction, is e(|uidistant from B and (7. It may also be proved that no point out of MB is equi- distant from B and C. The line MN is called the Locus of all the points, infinite in number, which are equidistant from B and 0. Def. In plane Geometry Locus is the name given to a line, straight or curved, all of whose points satisfy a certain geometrical condition (or have a common property), to the exclusion of all other points. ro4 E UCLID'S 1: A EMKXTS. \ Books I. & II. Next, suppose we luive to detenuine the position of a point, which is eqiiitli.stant from three given poiuLs A, L, 6', not in the same straight line. If we join A and B, we know that all points equidistant from A and B lie in the line FI), which bisects AB at right angles. If we join B and C, we know that all points equidistant from B and C lie in the line QE, which bisects BC at right angles. ij Hence 0, the point of intersection of PD and QE^ is the only point equidistant from A, B and C. PD is the Locus of points equidistant from A and B, QE i^'andO, and the Intersection of these Loci determines the point, wdiich is equidistant from .1, B and C. Examples of Loci. Find the loci of (1) Points at a given distance from a given point. (2) Points at a given distance from a given straight line. (3) The middle points of straight lines drawn from a given point to a given straight line. (4) Points equidistant from the arms of an angle. (5) Points equidistant from u given circle. (6) Points equally distant from two straight lines which intersect. [. & II. Books I. & 11.] SOLUTION OF PROBLEMS. 105 point, iiol in istant i right iistiint ; right is the Joint, mi a vhich Note 9. 0)i the Methods employed in the solution of - Problems, In the solution of Geometrical Exercises, certain methods maybe applied with success to particulai classes of questions. We propose to make a few remarks on these methods, so far as they are applicable to the first two books of Euclid's Elements. 3%e Method of Synthesis. In the Exercises, attached to the Propositions in the pre- ceding pages, the construction of the diagram, necessary for the solution of each question, has usually been fully described, or sufficiently suggested. The student has in most cases been required simply to apply the geometrical fact, proved in the Proposition preceding the exercise, in order to arrive at the conclusion demanded in the question. This way of proceeding is called Synthesis ((rii/^6o-ty = com- position), because in it we proceed by a regular chain of reason- ing from what is given to what is sought. This being the method employed by Euclid throughout the Elements, we have no need to exemplify it here. The Method of Analysis. .The solution of many Problems is rendered more easy by supposing the problem solved and the diagra/m constructed. It is then often possible to observe relations between lines, angles and figures in the diagram, which are suggestive of the steps by which the necessary construction might have been efiected. This is called the Method of Analysis (01^0X^0-1^= resolution). It is a method of discovering truth by reasoning concerning things unknown or propositions merely supposed, as if the one were given or the other were really true. The process can best be explained by the following examples. Our first example of the Analytical process shall be the 31st Proposition of Euclid's First Book. m io6 EUCLID'S ELEMENTS. [Books I. & ir Ex. 1. To dnuv a straight line thronyh a given point parallel to a given straight line. Let A be the given point, and BC be the given straight iine. Suppose the problem to be effected, and EF to be tha straight line required. E ^A. B O 77 Now we know that any straight liv^e AD drawn from A tf> meet BC makes equal angles with EF and BC. (i. 29.) This is a fact from which we can work backward, and arrive at the steps necessary for the solution of the problem ; thus : fake any point I) in BC, join AD, make z EAD= l ADC, and produce ^J. to F: then EF must be parallel to BC. Ex. 2. To inscribe in a triangle o. rhombus, having one of ita angles coincident with an angle of the triangle. Let ABC be the given triangle. Suppose the problem to be effected, and DBFE to be the rhombus. Then if EB be joined, l DBE= l FBE. This is a fact from which we can work backward, and deduce the necessary construction ; thus : Bisect z ABC by the straight line BE, meeting AC in E. Draw ED and EF parallel to BC and AB respectively. Then DBFE is the rhombus required. (See Ex. 4, p. 59.) \ Eooks I. & 11 J SOLUTION OF rRODLEMS. T07 the L\. 3. To (hlnrmine the loiut in a given draujJd line, at icJiidh straiyht lines, drawn from two rjiven joints, on the name xlJc of the (jiven line, make equal angles ivlth it. Le; CD he the given line, and A imd B the given ^joinlu. Suppose ihe problem to be effected, and P to be the point requii'ed. D We then reason thus : If BP were produced to some point A , I CPA', being- z BPD, will be= z APC. Again, if PA' be niiide equal to PA, A A' will be bisected bv OP at rifjht angles. Thi« is a fact from which we can work backward, and find the steps necessary for the solution of the problem ; thus : From A draw AG ± to CD. Produce ^0 to A', making OA':^OA. Join BA', cutting CD in P. Then P is the point required. Note It). On Sijmmetry. The problem, which we have just been considering, suggests the following remarks : If two points, A and A', be so situated with respect to a straight line (^D, that CD bisects at rijiht angles the straight line joining A and A\ then A and A' are said to he sy^nmetrical with regard to CD. The importance of symmetrical relations, as suggestive of methods for the solution of problems, cannot be fully shewn io8 EUCLID'S ELEMENTS. [Books I. & II to a learner, who is unacquainted with the properties of the circle. The followincr example, however, will illustrate this part of the subject sufficiently for our purpose at present. Find a point in a given straight line, snch that the sum of its distances from two fixed points on the same side of the Urn is a minimum., that is, less than the sum of the distances of any other point in the line from the fixed points. TakinjT the diagram of the last example, suppose CD to be the given line, and A, B the given points. Now if A and A' be symmetrical with respe«t to CD, we know that every point in CD is equally distant from A and A'. (See Note 8, p. 103.) Hence the sum of the distances of any point in CD from A and B is equal to the sum of the distances of that point from A' and B. But the sum of the distances of a point in CD from A' and B is the least possible when it lies in the straight line joining A' and B. Hence the point P, determined as in the last example, is the point required. Note. Propositions ix., x., xi., xii. of Book I. give good examples of symmetrical constructions. Note 11. EuclidJs Proof of 1, 5. The angles at the base of an isosceles triangle are equal to one another ; and if the equal sides be produced, the angles upon the other side of the ha^e shall be equal. Let ABC he an isosceles a, having AB^ 4C Produce AB, AC to D and E. Then must l ABC^ l ACB, and L DBC^ l ECB. Books I. £c II.] EUCLIUS PROOF OF /. 5. TOO 111 BD take any pt. F. From AE cut oft' AG=AF. Join FC and GB. to one oon the Then in as ^i^C, AGB, V FA==GA, and A(J=AB, and z i^^C= z GAB, .'. FC=GB, and z AFC= z JG^^, and z JCF^ z J.5(?. 1.4. Again, :' AF=AG, of which the parts ylJ5, J.Oare equal, .*. remainder JBJ'^= remainder CG. Then in as BFC, CGB, •/ BF=^CG, and FC=GB, and z JSi^'C- z C'6^B. .-. z FBC= z G^C^, and z BCF= z C£rA Now it has been proved that z ACF= z ^jB(r, of which the parts z JBCF and z CJBG^ are equal ; .*. remainmg z ^(7B= remaining z ABC. Also it has been proved that z FBC= z G^C'B, that is, z D5C= z ^Oi?. Ax. 3. 1.4. Ax. 3. ^.1 ♦ i Q. E. D. no r.UCrJD'S ELK.'^rEA /is*. [BooUb I. & II Note 12. EucHiTit Proof of I. 6. . If two angles of a triangle he pq'^inl to one another^ the Kide* aho, which subtend the equal anylen, shall be eq^uol to one another. Tn AABCht l ACB-~= l ABC. Thenmwt AB=^AG. For if not, AB is either p^roater or less than Ad Suppose AB to be jrreater than AO. From AB cut off BD=ACy and join DC. Tbenin AsDL'C, JC'B, •.• DB^AO^ and BC is common, and z DBC= i ACB, .'. aDBC=aACB; I. 4. that is, the less = the greater ; which is absurd. .*. AB is not greater than AC. Similarly it may be shewn that AB is not less than A C ; .-. AB=Aa Q. E. D. Note 13. Euclid's Proof of I. 7. Upon the same hose and on the same side of it, there cannot he two triangles that have their sides which are ter- minatcA in one extremity of the base equal to one another, and their sides which are terminated in the other extremity of the base equal also. If it be possible, on the same base AB, and on the same side of it, let there be two a s ACB, ADB, such that AC=AD, and also 5C-5D. Join CD. .sid( be It BooitB I. & yi.'j i:('(i ijys j'Muor of j. 7. MI I. 4. i*'ir^t, wlioh tlif veitox of each of the as is imUide th othfcr A (Fig. 1.) ; le Fio 1. ' -4 ••• AD=AC, .'. L ACDr^ . ADC. I. 5. But A ACD is rrveater thiiu l BCD ; . . i J.i>C' is yreuter than z ^(7D ; much more ia z BDC grenter tlian z £0i). Atrain, •.• BC^BJJ, .: L BI)C= u BCD, that is, z 7?i>(7 is both equal to and greater than z BCD ; A'hich is absurd. Secondly, when the vertex D of one of the as falls witJiin the other a (Fig, 2) ; Produce AC and AD to JEJ and F Then •.•.4e=JD. .-. lECD= iFDC. 1.5. But z iJCl> is greater than z X'CD ; .-. z i^/>6^ is greater than z J5CD ; much uiore is z jBZX* greater than z £(7D. Again, •.• BC=BD. .-. lBDC= I BCD; that is, z ^Z>C* is both tqual to and greater than z BCD : which is absurd. Lcistly, Avhen the vertex D of one of the as falls on a side BC of the other, it is plain that BC and BD cannot be equal. q. e. ^^ list EUCUiys ELEMENTS. [Booi^B I- ^ "• Note 14. EaclixTs Proof of L 8. // two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their bases equal, the angle which is containid by the two siiks of the one must be equal to the angle contained by the two sides of the other. Let the sides of the as ABC, DEF be fqual, each to each, that \%,AB= DE, A C=I)F a nd BC=' EF. Then mud l BAC=> l EJJF. it Apply the l ABC to the a DEF. so that pt. B is on pt. E, and BC on EF. Then '.' BC=EF, .'. C will coincide with F, nnd BC will coincide with EF. Then ylB and AC must coincide with DE and DF. For if AB and AC have a different position, as GE, OF, then upon the same bas-t and upon the same side of it there can be two as, which have their sides which are terminated in one extremity of the base equal, and their sides which are ter- minated in the other extremity of the base also equal : which is impossible. I. 7. .•. since base BC coincides with base EF^ AB must coincide with DE, and AC with DF ; .-. I BAC coincides with and is equal to z EDF. q. E. D. 11. Books I. & xi.j Axurmiji j'/w >/ oj' i. 24. 113 \) tiro banes of the les of ,0 each, E, GF, it there luited in are ter- which 1.7. XoTli io. Another Froof of 1. i4. In the AS Anc, Dill', let AD^-DE aixl AC^JjF.mA let L L M' be ^Uiitt^r thtm / EDF. Then ,itu»t JJC be (jixalti tlmu EF, . E. D. Apply the a DEF ii) the b. ABC so that DE coiiieide.s nith AB. Then •.• L EDF is less tli;iu i HAC, DF wiH fall heiweeu BA luul A(\ and i*' will lull on, ov above, or below, BC, 1. If F fall (.11 i^C, iJi' is less than !?(; ; .•. EF is leiss than BC. II. If i^ fall «/>ar. i?C', ijfjP, i'.4 together are less than BC, CA, mAFA^CA ; .-. BF is k'ss than BC ; .: EF IS less than BC. III. \iFM\ below BC. let Ji'^eitt ^Cin 0. Then BO, OF together are greater than BF, I. 20. and 0(7. AO \ AC; I. 20. .-. i?C, JLi^ i>T, .^C together, anil AF=AC, .• BC IS greater tl)an I?i^ • and .-. EF U less than BC. ^. K. D. h. 114 EUCLIJys Kl.llMKNTS. : Books I. & II. I I NoTK IG. iyaclUI''H Froof of 1. £G. If two iriaiKjles have two aufjhs of the one, equal to tuo angles of the oi'iitr, each to tach, and one side equal to one side, viz., either the sides adjacent to the equal ancjles, or the sides opposite to equal angles in each; then shall the other sides be equal, each to each ; and also the third angle of the one to the third angle of the other. a B In A s ABC, DBF, Let L ABC = L DEF, and / ACB = „ DFE ; and first, Let the sides adjacent to the equal z s in each he equal, that is, let BC=FF. Then must AB=DE,ii\^d AC=JjF,imd l BAC == i EDF. For if AB be not = DE, one uf them must be the greater. Let AB be the greater, and make C D^ DE, and join GC Then in lh GBC, DEF, •/ GB^DE, and BC=EF, and / GBC = l DEF, .'. L GCB= . DFE. I. 4. But z A CB -= L DFE by hypothesis ; .-. ^GCB= ^ACB- that is, the less = the greater, ^vhieh is impossible. .'. AB is not nreatcr than DE. In the same way it may be shewn that AB is not less than DE; .'. AB^DE. Then in as ABC, DEF, •: AB=DE, and BC EF, and z ABC= l DEF, .: AC=DF, and z BAC= . EDF. I. 4. 1.4. tooy.H I. & 11/ EUCLID'S J'ROUF OF I. 26. n; Nc:^X, let ti,e sides which are opposite to equal angles m each triangle be equal, viz., AB=DE. Theu, must AC=-.BF, and BC^EF, and i BAC ^ i EDF. A ■X B H C \ — ^ Forif i?C be not-^i^; let BC be the greater, and make im=EF, and join .4 /f. Then in A s A BH, DEF, \- AB=DE, and BE=EF, and . ABH=^ l DEF, .-. z AHB--= L DFE. I. 4. But L ACB - L DFE, by hypothesis, .-. ^AIIB = ^ACB; that is, the exterior z of a J.//6' is equal to the interior and opposite z ACB, which is impossible. .'. BC IS not greater than EF. In the same way it may be shewn that BC is not less than EF ; .'. BC=EF. Then ni a s ABC, DEF, V AB-^DE, and BC=EF, and z J /iC= z DEF, . . AC-^DF, and z iJ^C- z J^i^i?'. I. 4. Q. E. D. I. 4. ■ABH (i6 EUCLID'S Kf.KMENTS. [Books L & II. Miscellaneous Exerciser on Bools I. and II. n ■ ij, 1. A B and CD are equal sttaiffht line'*, bisoctinji oue another at right angles. Shew that ACBD is a square. 2. From a point in the side of a parallelogram draw a line dividing the pnrallelogram into two equal partK. 3. In the triangle FDC, if F(JD he a right angle, and angle FUG be double of angle CFD, shew that FD is double of DC. 4. \i ABC\)^ an equilateral triangle, and AD, BE be per- pendiculars to the opposite sides intersecting in F ; shew that the square on AB is equal to three times the square on AF. 5. Describe a rhombus, wiiich shall be equal to a given triangle, and have each of its sides equal to one side of the triangle. 6. From a given point, outside a given straight line, draw a line making with the given line an angle efpiul to a given rectilineal angle. 7. If two straight lines be drawn from two given points lo meet in a given straight line, shew that the sum of these lines is the least possibh', when they make e(jual angles with the given line. 8. A BCD 19, 'A parallelogiam, whosti diagonals AC, BD in- tersect in 0; show that if the parallelograms AOBP, DOCQ be completed, the straight line joining /* .ind Q passes through 0. 9. -4/>CZ), .£J7iC'i'^ are two parallelograms on the same base BC, and so situated that CF passes through A. Join DF, and produce it to meet BE produced in K ; join FB, and prove tha,t the triangle FAB equals the triangle FEK. 10. The alternate sides of a polygon are produced to meet ; shew that all the angles at their points of intersection together with four right angles are equal to all the interior angles of the polygon. 11. Shew that tne perimeter of a rectangle is always greater than that of the square equal to the rectangle. I. «6 II. nooks I. ec 11.] Mlsr.lU.LAM'OUS JiXKRClSES. w another IT a lino 111 !m;4le doiO)lo be pei- lew tliiio ^^Ah\ a given e of the ne, draw I a given joints to fse lines with the BD in- , Docq tliroii;j;^i line base oin m\ FBy and "to meet ; together ingles of 12. hliinv tluit tI;o opposite sides of an equiangular liexagon are parallel, thougli tlioy be not equal. 13. It* two equal straiglit lines intersect each other anywheic itt liuht angles, shew that the area of the quadi'i lateral fortiuil by joining their extremities is invariable, and equal to onn-hult I he square on either line. 14. Two triangles ACB, ADB are constructed on the tame side of the same base AB. Shew that if AC-BI> ami AD=nt\ then CD is parallel to AB ; but if ^."=i>(7and AD~BJJ, then CD is perpendicular to AB. 15. ^7i is the hj'potenuse of a right-angled triangle ^J5(7 : Hnd a point D in AB, such that I)B may be equal to the per- pendicular from D on AG. 16. Find the locus of the vertices of triangles of equal area on the same base, and on tin sniue j^ide of it. 17. Shew that the perimeter of an isosceles triangle is less than that of any triangle of equcJ area on the same base. 18. If each of the equal angles of an isosceles triangle be eqiuU to one-fourth the veitical angle, and from one of them a perpendicular be drawn to tlie base, meeting the opposite side produced, then wdltho part produced, the perpendicular, and the remaining side, form an eqnilateral triangle. 19. If a straight line terminated by the sides of a trianglo be bisected, shew that no other line tenninated by the same two sides c;m be bisected in the same jioint. 20. Shew how to bisect a given quadrilateral by a straigiit line drawn from one of its angles. 21. Given the lengths of the tv/o diagonals of a i-hombus, con- struct it. 22. ABCJJ is a (juadrilateral figure : construct a triangle whose base shall be in the lino A J), such that its altitude shall be equal to a given line, and its area equal to that of the quadrilatetal. 23. If from any point in the base of an isosceles triangle perpendiculars be drawn to the sides, their sum will be equal Meular from either extremity of the b. the perpt upon o ppo; site side. Is greater riH E(fCrrD'S ELEMENTS. [Books 1. & it 24. li AlMj be a trianolf, in which C is a right angle, and DE be drawn from a point D in ^Outright imj^les to AB^ prove that the rectanj^Mos ylL', AE and JC, AD are equal. 25. A line is drawn bisecting paralleiogram A.BCD, and meeting AI), BG in E and F : sliew that the triangles EBF, CED are equal. 26. Upon the hypotenuse BO and the sides CA, AB of a right-angled triangle ABC, squares BDEG, AF and AG are described : shew that the squares on DG and EF are together equal to five times the square on BG. 27. If from the vertical angle of a triangle three straight lines be drawn, one bisecting the angle, the second bisecting the base, and the third perpendicular to the base, shew that the first lies, both in position and magnitude, between the other two. 28. If ABG be a triangle, whose angle ^ is a right angle, and BE, GF be drawn bisstcting the opposite sides respectively, shew that four times the sum of the .squares on BE and GF is equal to five times the square on BG. 29. Let AGB, ADB be two right-angled triangles having a common hypotenuse AB. Join (JJJ and on 67.> produced both ways draw perpendiculars AE, BF. Shew that the sum of the squares on GE and GF is equal to the sum of the squares on BE and I)F. 30. In the base AG of a triangle take any point D: bisect AD, DG, AB, BG at the points E, F, G, II respectively. Shew that EG is equal and parallel to FII. 31. If AD be drawn from the vertex of an isosceles triangle ABG to a point D in the base, shew that the rectangle BD, DG is equal to the diiierence between the squares on AB and AD. 32. If in the sides of a square four points be taken at equal distances from the four angular points taken in oider, the figure contained by the straight lines, which join them, shall also be a square. 33. If the sides of an equilateral and equiangular pentagon be produced to iiieet, slicw tliat the sum (jf the angles at the points of meeting is equal to two ri^ht angles. Pl fi k if r. alt le, and to A C, lul. D, and , EBF, B of a AG are lOgether straight tisecting lew that eeu the It angle, ectively, id CF is i having roduced the 6ium squares : bisect jectively. triangle \BD, DO ind AD. at equal |der, the mi, shall lentagon Is at the Books I. & II. I MISCEL[..\Xi:OUS EX EliClSES. 119 34. De-;cril)e a S'juuie that shall be equal tu the ditlerence between two given and unequal squares. 35. ABQl). AEOF are two piirullelograms, EA, AD being in a straight line. Let FG, drawn parallel to AC, meet ^.I produced in G. Then the triangle ABE equals the triangle ADG. 36. From A C, the diagonal of a square A BCD, cut oft' AE equal to one-fourth o( AC, and join BE, DE. Shew that the figure BADE is equal to twice the square on AE. 37. If ABC be a, triangle, with the angles at B and C each double of the an^le at ^l, prove that the square on AB is equal to iho square on B(J together with the rectangle AB, BG. 38. If two sides of a quadrilateral be parallel, the triangle contained by either of the other sides and the two straight lines drawn from its extremities to the middle point of the opposite side is hall' the quadrilateral. 39. Describe a para lleloj^ ram equal to and equiangular with a given parallelogram, and having a given altitude. 40. If the sides of a triangle taken in order be produced to twice their original lengths, and the outer extiemities be joined, the triangle so formed will be seven times the original triangle. 41. If one of the acute angles of a riijlit-an^led isosceles, triangle be bisected, the opposite side will be divided by the bisecting line into two i)arls, such that the square on one will be double of the square on the other. 42. ABC is a triangle, right-angled at B, and BD is drawn perpendicular to the base, and is produced to E until ECB is a right angle ; prove that the sfjuare on BC is equal to the .sinn of the rectangles AD, DC and BD, DE. 43. Shew that the sum of the S(inares on two unequal lines is' greater than twice the rectangle contained by the lines. 44. From a given isosceles triangle cut oft" a trapezium, having the bas«^ of the t;iangle for one of its parallel sides, and having the other three side? tqutd. 120 R UCL m'S EL EMFXTS. [Books I. & II. 4/5. If any imiiiber of paiallelofirams Vi« constructed having thoir sides of given length, shew that the snin of the squares f^w the diacrnnals of ench will be the same. 4(!. ABCD is a right-angled piirallelogram, and AB is double of JiC ; on AB an equilateral triangle is constructed: shew that its area will be less tli.'in that of the parallelogram. 47. A point is tiilcen within a triangle ABC, such that the angles BOC\ CO A, ACB are equal ; prove that the squares on BCj CA, AB !ire together equal to the rectangles contained by OB, OC; (K\ 0A\ OAy OH-, and twice the sum of the squares on OA, OH, OC. 48. If the sides of an equilateral iiud equiangular hexagon be produced to meet, the angles formed by these lines are together equal to four right angle=5. 49. ABC is a triangle lighr-nngled at A ; in the hypote- nuse two points D, E are taken such that BD = BA nnil CE--CA ; shew that the square on DE is equal to twice tie rectangle contained by BE, CD. T)(). Given one side of a rectargie which is equal in area to i given square, Prd th^ n*l->p" ndc. 51. AB, AC ^xe the two eq-al sides of an isosceles trianglo : from B, Bl) is drawn perpendicular to AC, meeting it^in V; shew that the square on BB is greater than the square on Of> by twice the rectangle AD, CD. 2. 3. I 5. 6. 7. 8. APPENDIX. EXAMINATION PAPERS IN EUCLID SET TO CANPIDATES FOR First and Second Glass Provincial Certificates, ANn TO f«TUT>KNTS MATRICULATING IN THE UNIVERSITY OP TORONTO. II t0t ■ SECOND CLASS PROVINCIAL CERTIFICATES, 1871. TIME — TWO HOURS AND A HALF. 1. If two *riangles have two sidea of the one equal to two sides of the other, each to each, and have likewise tlieir bases equal, the angle which is contained by the two sides of the one shall be equal to the angle contained by the two sides, equal to thf^m, of the other. 2. ' Triangles upon the same base, and between the same par- allels, are equal to one another. 3. If the square described upon one of the sides of a triangle be equal to the squares described upon the other two sides of it, the angle contained by these two sides is a right angle. 4. If a straight line be divided into two equal, and also into two unequal, parts, the squares on the two unequal parts are together double of the square on half the line, and of the square on the line between the points of section. 5. If a straight line be divided into any two parts, the rec- tangles contained by the whole and each of the parti=i are together equal to the square on the whole line. 6. Bisect a parallelogram by a straight line drawn from a point in one of its sides. 7. Let A B C be a triangle, and let B D be t» straight line , drawn to D, a point in A C between A and C, then, if A B be greater than A C, the excess of A B above A is less than that (,f B D above D C. 8. In a triangle A B C, A D being drawn perpendicular to the straight line B D which bisects the angle B, show that a line drawn from T> parallel to B C will bisect A C. WoTE. — The percentage of marks requisite, in order that a cajididate may be ranked of a particular grade, will be taken on the vaJue of the above paper, omitting question 8. !! U. APPENDIX. SECOND CLASS PROVINCIAL CEKTIFICATES, 1872.. TIMR— 2| HOURa. 1. Define a iftrnifjht linr, a /*/«?/»' rechhneal anrjte, ariijht anplc^ ■. a Gnomov. Eimuciuto Enrlid's Postulates. 2. If from the ends of the Bide of a tviiiugle there be drawn two straight lines to a point within the triniif^le, tljose shall be less tlian the other two sides of the triangle, but shall contain a greater anj,'le. 8. If two triangles have two auf^'b^i? of the one equal to angle? of the other, each to ei\ch, and one side equal to one side, namely, either the sides adjacent to the equiil angles. line between the points of seotiou, is equal to, &c. (5, II.) In every triangle, the sijuare on the side subtending an acute angle is less than the sqiiarcs on the sides con- tfiining that angle, by (fee. (13,11). (It will be suf- lieent to take the case in which the perpendicular falls within tlie triangle.) To describe a square tliat shall bo equal to a given rncti- lineal figure. The squa e o)i ai ; .straight lino drawn from the vertex of an isosceles triangle to the base is less than the square on a side of a triangle by a rectangle contained by the segments of the base. 9. 10. 11. SECOND CLASS PROVINCIAL CEETIFICATES, 1875. TIMK — TWO HOUKS AND THREK-QUARTEKS. Note. — Those students who take only Book I. will confine themselves to the first seven questions. Those who take Books I. and II. will omit the questions marked with an asterisk (*), namely, (1) and (2). *1. If one side of a triangle be produced, the exterior angle is greater than either of the interior oppo.site angles. *2. If two triangles have two angles of the one equal to two angles of the other, each to each, and oiie side equal to one «ide, namely, the sides opposite to equal angles, then shall the other sides be equal, each to each. 3. If a straight line falling on two other straight lines make the alternate angles equal to each other, these two straight lines shall be parallel. MPVfnriMX. uideH hto jfVit hiio8, Iher, the ther. the same ncl on tht nun their it aiifiles. is a ripht and join twice thf s and also tallied by in tht) lino 5. (5,11.) ending an sides con- 11 be suf- icular fiiUfc! iven rncli- vertex of than the contained I, L876. ill confine who take with an irior angle ite angles, ual to t\s'o side equal lal angles, )ach. ines make these two 6. 6. 7. 8. <0. If a ^trniglit (ino fall upon ivro prirallcT Ptrmfr/'T nrf, it nuikd triangle, if a perpendicular be drawn fntiii llu' lif^lit an^Ie to the b;ise, the liiiuigles on each Fiile (if it are similar to the whole Iriiinjjle and to one aii'it her. 6* II U>ur struighf linos bo piv^ji Mtioiial":, tho similar rectilinoal li^jiiies ciiiii.arly dtsciihcd ujioij them bh.iil uiau bo pr.ipinti>iii;ils. C. Dr:»w a strtii;,'lit lino po afs to touch two given circles. 7. Let A 15 C I e a triaii;::e. and from 1> and C. the e.xtrcmi. ties of InC *la^.. •' C. let iiiie ii F nnd (' K be drawn to V and £, Lho middlu points uf A C and A B rtspect- APl'ENDIZ. ively, then, if B F := G E, A B and A C shall be equal to one another. 8. DeBoribe an e(iuilatural triangle equal to a given triangle. 1. 2. 8. 4. 6. 6. 7. 8. 1. 2. 3. 4. 6. 6. FIRST CLASS PROVINCIAL CERTIFICATES, 1872. TIME — TWO AND A HALF UOOKS. If a Btruight line touch u circle, and from the point of con- tact a straight line be drawn cutting tUu circle, the angles which this line makes with the line touching the circle uhall be equal to the angleb which are in the alternate Begments uf the circle. To inbcribe a circle in a given triangle. Equal triangles which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional. Similar triangles are to one another in the duplicate ratio of their homologous sides. In any right angled triangle, any rectilineal figure described on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle. Two circles cut each other, and through the points of sec- tion are drawn two paralk-1 lines, terminated by the circumferences. Prove that these lines are equal. Let A C and B D, the diagonulH of a quadrilateral figure A B C D, intersect in E. Then, if A B bo parallel to C D, the circles described about the triangles ABE and C D E shall touch one another. Divide a triangle into two equal parts by a straight line «t right angles to one of the sides. FIRST CLASS PROVINCIAL CERTIFICATES, 1878. TIME — THREE HOURS. The angle in a semicircle is a right angle. A segment of a circle being given, describe the circle f>i which it is a segment. Give Euclid's definition of proportion ; and prove, by taking equi-multiples according to the definition, that 2, 3, 9, 13, are not proportionals. Similar triangles are to one another in the duplicate ratio of their homologous sides. To find a mean proportional between two given straight lines. Through C, the vertex of a triangle A C B, which has the Bides A C and C B equal to one another, a line C D 8 1. 2. B. A. 6. 6. 7. eeqaal angle. 872. of con- rcle, the oucUiug ■u in the ^qual to tout the ite ratio lescribed al to the ho bides B of BCO- id by the uual. U figure rallel to it line «i 1878. circle '^f )y taking t ^, Of «/i ate ratio straight has the line C D ▲PPKNDIX. vii. is drawn parallel to A B ; and straight linog, A D, D li, arn driiwu from A aiul II to any jxtint D in C D. Provo that tlio unglo A C D in gruiitor th;ui tho angle A I) B. 7. A 15 C 1) is a (iiiadrilateral ti;,'are iuBoribed in a circio. From A ami 15, poipondiiuilurs A E, 13 F are let fall on CD (|)roilucoi.i if ixcoHsiiry) ; lunl fruin C mid D, porpoiuliculuiB C (Jr, i) 11, uro lot fml on 15 A ^^)ro(lU(•ell if nooossiiry). Provo that tho rontmiglorf A E, B F and C (>, D II, aro ciiual to one anotlior. 8. A B C D in a quadrilatoriil tiguro inscriljcd in a circle. The straight lino D E drawn through D piirallel to A B, cuts tho Hide B C in E ; and tho straight lino A E pro- duced moots D C produced in F. Prove, that if tli« rectangle B A, A I) be equal to the rectangle E C, C F, tljo trianglo A D F hhall be ociuul to the quadrilateral ABC D. 1. B. 4. 6. 6. 7. FIRST CLASS PROVINCIAL CEUTIFI0ATE3, 1874. TIME — TUUKE H0UK8. In equal circles, equal straight lines cut off equal ciroum- ferenced, the greater, equal to tho greater, and the less to tho less. To describe a circle about a given equilateral and equiangu- lar pentagon. To find a moan proportional between two given straight Hues. What is meant by duplicate ratio ? Write down two wliole numbers, which arc in the duplicate ratio of J to ^. What aro similar rectiUneal figures ? Similar triauglos aro to one another in the duplicate ratio of their iiuuiolugous sides. In itny right anglod triangle, any rectilineal figure described on the sido suotending the right angle is equal to the eimilar and similarly described figures ou the sides containing the right an>,'lo. To describe a trianglo, of which the base, the vertical angle, and tho sum of tho two sides aro given. From A the vertex of a triangle ABO, in which each of the angles ABC and \CBis less than right angle, AD is let fall perpeudicular on the base BC. Produce BC to E, making CE equal to AD ; and let F be a point in AC, such that the triangle BFE is equal to the tri- angle ABO. Provo that F is one of the angular points of a square inscribed in tho triangle ABC, with one of its sides on BO. ■M0m -ft f! viii. AmnrDix. 8. Let E be the pofnt of interFPrtion of the dinponnls of a quadrilatcrixl fipure AB(,'I), of wbicli any two opposite Biifflos uro together e)» = a'^ -f 2(ilt 4. b'^ . Of all the squares that can be inscribed within an- other the least is that formed by jjiniug the bisec- tions of the side. 6. If a strai^'ht line be divided into two eqnal and also into two UMefpial parts, the sqttares on the two unequal parts are tvith what limitutiun this can be done. MMHr t. APPENDIX. 10. "When is a drele said to be inBcribed in a rectilme, 7. If a straight line touch a circle, the strai^nt. Hfw lnt the converse uscribed angles, it angle. 10 au^lea 11. 12. A tangent to a circle is drawn at an angular point of au inscribed regular pontiigon, and a side jn-oduceu tl\rou;,'h that paint, show tliit a straight line making equal intercepts on the tangent and the side produced, is parallel to the tangent at ouo of the adjacent an;ju- lar p jints. To describe a circle about a given equilateral pentagon. With an angular point of the regular pentagon as centre, and a side as radius, describe a second circle ; show that the tangent to the first circle at a point of inter- section of the circles meets the common diameter at a point without the second circle. In the above show that the distance from the above point to the centre of the first circle is greater than the diameter of the second circle. MATillCULATION, 1874. noNons. •,• Nos. 1 and 3 to bo omitted for Senior Matriculation ; Nos. 12 and 13 to bo omitted for Junior Matriculation. 1. Parallelograms upon the same base and between the same parallels are equal to one another. From the centre O of a circle the radii 0/f, Oli are drawn, the tangents at .1 and li meet in C \ if 00 be bisected in D and l)E be drawn perpendicular to-OL* meeting OB in E, then AE will bisect the figure one A. 2. In every triangle the square on the side subtendiagany of the acute angles is less than the squares on the .siiles containing that angle by twice the rectangle eoutaineJ by either of these sides, and the straight line inter cei)ted between the perpendicular let fall upon it froi^ the opposite angle and the acute angle. Construct a square that shall be equal to the difference between the sum of the sipiares on two given straight lines and the rectangle under these lines. 3. Through a given point to draw a straight line parallel to a given str.d.;ht line. From a giren jjoint in the circumference of a circle to draw a chord, when possible, that shall be bisected by a given chord. 4. Find the sunj of (1) all the interior angles of any recti- Jineal figure ; (2) all the e.\terior angles. AB, CD the alternate sides of a regular jioiygon are pro- duced to meet in E, if AC, OE meet in F, U Iteing the centre of the polygon, show that AF.EC — OF.FV XIV. APPENDIX. 5. 6. 7. 8. 10. 11. 12. To divide a given straight line into two parts, so that tho rectangle contiiinetl by the whole and one of the parts shall be equal to the square on the other part. If AB be bisected in C and produced to a point D, such that AC.CD-^AD.DB, then AD is divided in C in the manner required by the proposition. If from any point without a circle two straight lines be drawn, one of which cuts the circle and the other touches it, the rectangle contained by the whole line that cuts the circle and the part of it without the circle shall be equal to the square on the line that touches it. Any number of circles pass through two given points A and B ; shew that with any given point C in AB pro- duced, as centre, a circle may be described cutting the other circles at right angles, and find its radius. To draw a straight line from a given point either without or in the circumference which shall touch a given circle. Find the point in the line joining the centres of two circles of different radii, such that if a perpendicular be drawn through it, the tangents to the circles from any point in this perpendicular may be equal. The angle at the centre of a circle is double of the angle at the circumference upon the same base, that is, upon the same part of the circumference. If a circle be described touching one of the equal sides of an isosceles triangle at the vertex and having tho other side as chord, the arc lying between the vertex and base is one-half the arc subtended by the chord. If a straight line touch a given circle and from the point of contact a straight line may be drawn cutting the circle, the angles made by this line with the line touching the circle shall be equal to the angles which are iu the alternate segments of the circle. To inscribe an equilateral and equiangular pentagon in a given circle. If two diagonals of a regular pentagon intersect and a circle be described about the triangle of which the greater segments are two sides, two sides of the pen- tagon which terminate at the other extremities of these segments are tangents to the circle at these points. To describe a circle about a given square. Find the relation between the areas of the oirclei described about and inscribed in a given square. If a straight line be parallel to the base of a triangle it will cut the sides, or the sides produced, proportion- ally, and if the sides, or the sides produced, be cat Al^ENDIX. tv. proportionally, the straipht line which joins the points of section shall be parallel to the base. 13, To find a mean proportional between two given straight lines. JUNIOR AND SENIOR MATRICULATION, 1875. * Tunior Matriculants will omit questions 15 and 16, and Senior Matriculants questions 12 and 13. 1. Define the terms axiom, postulate, scholium, corollory. 2. If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle con- tained by the two sides of the one greater than the angle contained by the two sides equal to them, of the other, the base of that which has the greater angle shall be greater than the base of the other. 3. If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles ; and the three interior angles of every triangle are together equal to two right angles. 4. Triangles on equal bases and between the same parallels are equal to one another. 6. If the square described on one of the sides of a triangle be equal to the squares described on the other two sides of it, the angle contained by these two sides is a right angle. 6. If the diagonals of a quadrilateral bisect each other, it is a parallelogram : if the bisecting lines are equal it is rectangular ; if the lines bisect at right angles it is equilateral. 7. If a straight line be divided into two equal, and also into two unequal parts, the squares on the two unequal parts are together double of the square on half the line and of the square on the line between the points of section. Divi(3e a straight line into two parts, so that the rectangle contained by the whole and one of the parts may be equal to the square on the other part. In the Algebraic solution of the preceding problem, we obtain a quadratic equation which gives two values of the unknown quantity. Enunciate the Geometrical proposition which corresponds to the other root. 10. The sura of the squares on the diagonals of a parallelo- gram is equal to the sum of the squares on the sides. 11. The opposite angles of a quadrilateral inscribed in a circle are together equal to two right angles. 12. The straight lii les bisecting the sides of a triangle at right angles meet in a point. 8. 9 TVi. APPEHOTS., 13. Construct a triangle, having given the tnidille points of Bidea. 14. Describe a circle about a given equilateral and equiangu- liir i)pntag(ui. 15. From a ^ivon Ktraight line to cut oiT any part required. 16. Similar trimiglcs are to ono auotlier iu tlio Juplicuto ratio 1 of their liomologous sides. 1 m SH ■ TIME — 3 nouns. 1. Describe an equilateral triangle upon a given finite straight Hue. By a method similar to that used in this problem, describe on a given finite straight line an isosceles triangle, the sides of which shall be each equal to twice the base. 2. li a straight line fall on two parallel straight lines, it makes the alternate angles equal to one anotht r, and the exterior angle equul to the interior and opposite angle on the same side ; and aloo the two interior angles on the same side togetlier equal to two right angles. What obj»'ctions have been urged ngninst the doctrine of ])arailel straight lines as it is laid down by Euclid? Where does the dilliculty origuiate and what has been isu>:gestcd to remove it ? 8. Iu any right angled'triangle, the squares described on the sides containing the right angle are together equal to the square of the side subtending the right angle. Show, by describing a scjuare on the outer side of one side, and on the inner side of the other, that the two squares thus described will cut into three pieces, so as exacily to make up the stpiare of the 'lypott-iiuse. 4. Divide alijehraicalUj a given line [a) into two parts, such that the rectangle contained by the whole and one part may be equal to the square of the other. Deduce Euclid's construction from one solution and explain the other. C. If two straight lines within a circle cut one another, the rectangle contained by the si'gnients of one of them is cqaal to the rectangle contained by the segments of the other. If, througli a point within a circle, two equal straight lines be drawn to the circumference^ and produced, they will bo at the same distance from the centie. 6. Explain and illustrate the fifth and seventh definitions in the fifth book of Euclid, and shew that a magnitude has a greater ratio to the less of two unequal maguitudea than it has to the greater. 7. With the four lines contain a-|-?>, a-\-c, a — &, a — c units respectively, construct a quadrilateral capable of hav- ing a circle inscribed iu it. ' APFICNDIX. xvfl. Piore tlmt no pm-nllplogram cnn be InRcrTbed In a circle excopt n )cotiu)},'lft ; and that no parallelogram can be (l^'scril)c{l about a circle ox(;ept n rlionili. B. Similar triungles are to one another in thu dnplicato ratio of their homologous sides. How iIopp if. Qpp( nr from Euclid that the duplicate ratio of two maguiiudes is the same as that of their squares ? lines they units Lav- FIRST CLASS PROVINCIAL CERTIFICATES, JULY, 1876 timp:— TiriiEE nouRB. N. li. — Alpehrnic symbola viust vot he used. I. (a) The straight line drawn at right angles to the diameter of a circle from the extremity of it, falls without the circle ; and no straight line can he drawn from the extremity, between that straight line and the circum- ference, so as not to cut the circle. (Ill IG.) ijb) Draw a connnon tangent to two given circles. How many can be drawn ? (Aiiollonius.) 9 (a) The oppoKite angles of any quadrilateral figure in- acribcd in n circle are together equal to two right angles. (HI 22.) (6) If straight lines be drawn from any point on the cir- cumference of a circle perpendicular to the sides of an inscribed triangle, their feet are in the same straight line. (il/. F. Janohi.) P (a) If the chord of a circle he divided into two segments by a point in the chord or in the chord produced, the rectangle contained hv these segments will bo equal to the dillercnce of the squares on the radius and on the line joining the given point within the centre of the circle. What propositions in Euclid follow immediate- ly from this ? (&) Describe a circle which shall pass through a given point and touch two straight lines given in position. (Al)oUo}:\m.) 4. (a) To dehcrihe an isosceles triangle, having each of the angles at the base d(nil)le of the third angle. (IV 10.) {h) Construct a triangh? having each i)f tlie an>»les at the baso e(pial to seven times the third angle B (a) If the veitieal a))gle of a tiianglo be bisected h? ft straight line which also cuts the base, the .segments of the l)ase have the same ratio whieli the other sides of the triangle have to one another ; and, if the segments of the base have the same ratio which the other sides of the triaugle have to one another, the straight lino ICVIU. APPENDIX. i I i drawn from the vertex to tk« poiut of Koction uhall bisect the vertical angle. (VI 3.) {b} The points in which the bisectors of the externaJ angles of a triangle meet the opposite sides, lit) in » ■traii^ai line. u 1 APPENDIX. sin. SECOND CLASS CEliTIilCATES, JULY. 187C. Tliia — rUUEK H0DB8. N-B. — Alijehraic symbols must not be used. Cundidatef wlto take Book II will omit QuaLioiis 1, 2 and 3, marked^. ValuM. I 16 3 16 6 16 . The angles at the base of au iKOBoeles triangle aro equal to one another ; and if the equal sides be produced, the angles on the other aide of the base shall bo equal to one another. Where docs Euclid require tho second part of this theorum? *2. If two triangles have two sides of tne one equal to two sides ol the oiiior, each to each, but the angle contained by two sides of one of them greater than the angle contained by the two siiles equal to them of the other, tho base of that which has the greater angle shall be greater than the base of the other. Why the restriction " Of the two sides DE, DF, let DE be the side which is not greater than the other"? •3. If two triangles have two angles of the one equal to two angles of the other, each to each, and have also the sides adjacent to the equal angles in each, equal to one auotlier, then shall the other side bo equal, each to each ; and also the third angle of the one to tho third angle of the other. (Prove by superposition.) What propositions in Book I aro thus proved ? 4. If a straight line fall upon two parallel straight lines, it makes the alternate angles equal to one another, and the exterior angle equal to tlie inte- rior and opposite angle on the same side; and also the two interior angles on the same side together equal to two ri^'ht angles. What objection may be taken to the twelfth axiom 7 What is its converse ? o. In any right-aiij^'led triangle, the square which ia described on the side subtending the right angle is equal to the squares described on the sides which contain the ri^dit angle. Prove also by dissoction and superposition. 6. Draw through a given point between two straight linos not parallel a straight line which shall be I bisected in that point. \H I 7. The perpendiculars from the angles of a triangle on the opposite uiiius meet in a point. 3 16 8 2 16 12 18 DC AFPUNDf'X. 20 8. Given the lengths of tho lines drnwu from the aiiKlos of a tiiiiiigle to tho points of biKOCtion of the opposite sick'S, construct tho triangle. 20 0. If a straight lino 1)0 cUviiloil into two parts, the eqnaro on tho whole line is cqunl to tho Rcjuares on tho parts, together with twice the rectangle C(nituinccl by the parts. 20 10. In every tiiungle, the siinare on tho side Bnbtending on acute angle is less than tho pcpuires on the Bides containing that angle by twice tho rectangle contained by cither of these eidcs, and tho straight line intercepted between the peipendicular let fall ou it from the opposite angle, and the acute angle. from the ectiou of 3. arts, the I fitjuares I'tiutaugle btcnding H on the rectangle 3 straight ar let fall Lto angle. ,ammifmsm»mt''