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Sir Wilftid Laurier, 18 x 24, price, post paid, 25 cents. , Sir John Macdonald, 18 x 24, price, post paid, 25 cents. Sir Oliver Mowat, 18 x 24, price, post paid, 25 cents. Magna Charta, 22 x 28, colors, 75c. Eng- lish Facsimile of Magna Charta, 22 x 28, plain, 50c. The two pictures for $1.00. THE EDUCATIONAL PUB. CO. 11 Richmond St. West, - Toronto Teachers Have you seen our Monthly Report eards ? These Cards are used in all the TORONTO SCHOOLS^ and are giving entire satisfaction. They are made of strong manilla card, and are GOOD FOR A YEAR. Sold at 25c, per dozen. Should be in every school. ^PRESS J -& The Edncational Pnblisbing Co., 1 1 Richmond St. W., TorQIito \ •-P ♦» w m uio 1,25 show- irince, )lorcd price, post post paid, Sng- 1x28, Dnto rt *m are /ery I' ■* PROPOSITIONS I. TO XXYI. PROPOSITION 1. /. If the two circumferinces intersect also at F, •what kind of* triangle will be found by joining AF and BF f AF = AB. Def. of a circle. BF = AB. • Def. of a circte. .:.AF=BF. Axiom I. , Thus AF, AB, and BF are all equal, and an equi- ' lateral triai^e ABF has been described on AB.. 2. What kind of quadrilateral is the fissure ACRF?' * See figure in No. i. Sine* AC, AF, BC, and BF each equal AB. Def. of a circle. .".AC, AF, BC, and BF' are all equal. Ax. i. That is ACBF is equilateral or a rhombus. Def. 25. PROPOSITION II. , . /. Under what circumstances would the i>oint D lie- (a) Outside th^circle CEF f [b) On the circumf^nce of the circle (^F? (a) The point D wHuld lie outside the circle CEF if the point A were farther from B than the length of BC. For then AB and BD, which is equal to it, would be longer than BC, the radius of the circle CEF. , (b) The point D would lie on the circumference of the circle CEF if the point A were the same dis- tance from B that the point B is from C. ,For then AB and BD, which is equal to it, woulcfbe radii of the cifcle CEF. a. If D were without the circle CEF, would it be necessary to produce DB ? It would not be necessary to produce DB. • BD-AD. Def. of an equilateral A DE = DG. n*f. of a circle. .•.BE = AG. Ax. 3. But BE*BC. -Def. cfacirciw .*.AG = BC. Ax. I. Wherefore from the point A a straight line AG has been drawn equal to BC. 3. Could the problem be solved by producing BD instead of DB ? Yes as in the figure — * B D = AD. Def. of an equilateral A' Det. of a circle. , Ax.^. Def. of a circle. * Ax. I. DE = DG .-.BE -AG. But BE = BC. .AG=BC Wherefore from the pojnt A a straight line AG has been drawn equal to BC. • 4. Could the problem be solved by joining AC instead of A B f Yes ; as in the figure — CD = AD. Def. of an equilateral A- DE = DG. .•.CE = AG. ButCE = CB. . .•.AG = CB. Def. of a circle. Ax. 3. Def. of a circle.* Ax. I. Wherefore from A a straight li^e AG has been drawn equal to BC. J. Does the line AG alwaff Re in^he sanit direction, no matter which of the 'above methods of construction is used? The line AG does not always Re in the canil direction, as may be seen from the diagrams. , • EUCLID A great many excellent exercises may be made irom this second proposition, (or you may : (a) Draw the A as in text-book, and then produce DB either down, as in the book, or up, as in question 3. (*) Draw the A below AB and produce DB each way. - . (c) Join A to C as in question 4, and make the A above AC, and produce DB either way. (rf) Join AC and make the A below AC, and produce DB either way. (e) Let these four be drawn, keeping thrt dis- tance from A to the end of BC, to which it is joined, shorter than the length of BC. Then change all by making this distance greater. This will give, in all, sixteen figures. Note. — You will have no difficulty with any of them if you remember the following : 1. The centre of the first circle is the end of the given line to which the point is joined. 2. The first side produced is that side of the triangle passing through the apex of the triangle and the centre of this circle. 3. The centre of the second circle is always the apex of the triangle. 4. The second side produced is that side which passes through the apex of the triangle and the given point. These are always true, no matter how the figure may be arranged. PROPOSITION III. I. AB is a gii>en straight line. Produce the line, making the whole length double that oJAB. Let AB be the given straight line. It is required to describe an isosceles triangle on AB, each of whose sides shall be double of AB. From A draw AC double of AB. Deduction t. From B draw BD double of AB. Deduction i. With centre A distance AC describe a circle. Post. 3. With centre B distance BD describe a circle. Post. 3. Let the circumferences intersect at E. Join AE and BE. Post. i. Then ABE shall be the triangle required. AE = AC. Def. of a circle. AC = 2AB. Cons. ;.AE = 2 AB. Ax. I. In the same manner it may be shown that BE = 2 AB. .•.AE = BE: Ax. I. Wherefore on the base AB an isosceles triangle ABE has been constructed, each of whose sides is double of AB. Let AB be the given straight line. It is required to produce AB, making the whole line thus produced double of AB. With centre B distance I3A describe a circle. Post. 3. Produce AB to meet this circle in C. Post. 2. Then AC shall be the line required. AB = BC. Def. of a circle. .•.BC = 2AB Wherefore the straight line AB has been pro- duced, so that the whole line AC thus produced is double of AB. s. Describe an isosceles triangle on a given straight line, such that each of the equal sides shall be twice as long as the base. 3. On a g'ven straight line describe an isosceles triangle having each of the equal sides equal to another given straight line. Let AB and C be the given straight lines. It is required to describe on AB an isosceles tri angle having each of its sides equal to C. From A draw AE = C. From B draw BF = C. With centre A distance Prop. 2. Prop. 2. AE describe a circle. Post. 3. With centre B distance BF describe a circle. Post. 3. Let the circumfwences intersect at D. Join AD and BD. Post. i. Thr.n ADB ahal! be the triangle required, AD =^ AE. Def. of a circle. AE = C. Cons. .•.AD = C. Ax. I. I i DEDUCTION In the same manner it may be shown that BD = C. .•.AD = BD. Ax. I. Wherefore on AB an isosceles triangle ADB has been constructed, havinfj^ each of its sides equal to the given straight line C. 4. Draw a straight line three times as long as a given straight line. This may be easily deduced from the first de- duction on this proposition. With C as centre and CB as distance, describe another circle, and produce AC to meet this circle in F. Then AF is the line required. 5. On a given straight line describe an isosceles triangle having each of the equal sides three times as long as the third side. This may be easily deduced from the second deduction on this proposition. From A draw AC three times as long as AB. Deduction 4. From B draw BD three times as long as AB. Deduction 4. Then proceed as in the solution of Deduction 2. 6. From a given point C, in a straight line A B, draw a straight line equal to AB. This is a par :ular case of the second proposi- tion, and if the iiints given at the end of the treat- ment of that proposition be remembered no diffi- culty will be experienced with this problem. DC = DE. Def. of a circle. DC = D B. Def. of an equilateral A- .•.CG = BE. Ax. 3. But BA = BE. Def. of a circle. .•.CG = BA. Ax. I. Wherefore from the point C a straight line CG has been drawn equal to the given straight line AB. 7. Produce the less of two given straight lines, making it equal to the greater. Let AB and C be the two given straight lines of which AB is the shorter. It is required to produce AB making it equal to C. From A draw AD = C. Prop. 3. With centre A distance AD describe a circle. Post. 3. Produce AB to meet the circumference in E. Post. 2. Then AE shall be the line required. AE = AD. Def. of a circle. AD = C. Cons. •.AE = C. Ax. I. Wherefore the straight line AB has b!en pro- duced to E making 'he whole line thus produced equal to C. PROPOSITION IV. 7. The sides of the square A BCD are equal to the sides of the square EFGH. Show that : {a) The diagonals AC and EG are equal, {b) The diagonals A C an^ BD are equal, {c) The diagonal AC bisects, that is, divides into two equal parts, the angle BAD. (d) The squares are equal in area. (a) In the A's ADC and EHG r AD = EH. \ DC = HG. Uadc=z.ehg. .•.AC = EG. {b\ In the A's ADC andDCB f AD = BC. \ DC-DC. Uadc=z.dcb. .•.AC = BD. {c) In the A's ADC and ABC AD = AB. DC = BC. ^ ADC =^ ABC. ^DAC= /.BAC. {d) In the A's ADC and EHG AD = EH. Hyp. DC = HG. Hyp. ^ADC=^.EHG. Def. of a squat e.- .-. A ADC -A EHG. Prop. 4- Similarly it may be shown that theAABC = AEFG. .•. the square ADCB = the square EHGF. Ax. 2. 2. A straight line AD bisects the vertical angle BAC of the isosceles triangle ABC, and meets the base at the point D. Shoiv that u is ine miasic point of the base, and that AD is perpendicular to EC. {. i Hyp. Hyp. Def. of a square. Prop. 4. Def. of a square. Common. Def. of a square. Prop. 4. Def. of a square. Def. of a square. Def. of a square. Prop. 4. m EUCLID 111 {, In the A's ADB and ADC AB = AC. Uef. of an equilateral A- AD = AD. Common, z. BAD =^ CAD. Hyp. _ BD = DC. Prop. 4. That is, D is the middle point of the base. Also z.ADB=iLADC Prop. 4. .'.AD is perpendicular to EC. Def. 7. J. If two straight lines bisect each other at right angles, any point in either of them is equidistant from the extremities of the other. Let AB and CD be the two given straight lines, bisecting each other at right angles in the point F. In CF take any point E. It is required to prove that E is equally distant from A and B. Join AE and BE. In the A'sEAF and EBF AF = BF. FE = FE. ^AFE=^BFB. .AE = BE. Post. I. r a, { Hyp. Common. Ax. II. Prop. 4. ^. The middle points of the sides of a square are joined in order. Show that the quadrilateral formed by these joining lines is equilateral. Let A BCD be the given square, and E, F, G, and H the middle points of the sides, and let EF, EG, G Hand HE be joined. It is required to prove EF = FG = GH = HE. In the A's AEH, BFE, CGF and DHG the sides are all equal. Def. of a square, and Ax. 7. The ^'s A,B, C and D are all equal. ^ Def. of a square, and Ax. 1 1. / .'jEF, FG, GH and HE are all equal. Prop. 4. .,„y,„_^^ ** ^^'"^'''^ E '^ '* point in A B, and rW^^^mVi Cif, '»''"" Ihai AE is equal to CF; EF is jefned^^J^^ equal to the angle CFE. th ea Let ABCD be the given square, E and F the * given points, and let EF be joined. It is required to prove that ^ AEF= ^CFE. Join AF and EC. Post. i. In the A's ADF and CBE 'DF = BE Ax. 3 AD = CB Def. of a square. ^ D = z. B Def. of a square, and Ax. 1 1. £.DAF=^ECB. Prop. 4. And AF = EC , Prop. 4. Now, since ^DAE= ^BCF Dff. of a square, and A::. 11.^ {' th th at at ba And ^DAF-^ECB ^FAE=^ECF Iiithe A's AEFandCEF • AE=CF. AF = EC. ^FAE=z.ECF. .-. iLAEF-=.LCFE. 1 iij {, Proved equal. Ax. 3. Hyp. Proved equal. Proved equal. Prop. 4. PROPOSITION V. /. Prove that the diagonal of a rhombus divides it into two isosceles triangles. th di Let ABCD be the given rhombus and AC oncof its diagonals. It is required to prove ABC and ADC are isos- celes A's. AB = BC. Def. of a rhombus. .'.ABC is an isosceles A. Again, AD = DC. Def. of a rhombus. 3. Prove that the opposite angles of a rJiombus are equal. E C DEDUCTION ' the .X. 7. i. II. »p. 4. ami C7-; •gual the ISt. I. Ix. 3 uate. X. II. Dp. 4. op. 4. s. II._ ;qual." .X. 3. Hyp. qual. :qual. 3p. 4. wides >ne«of : isos- nibus. nbin. imbus From the figure in deduction (') above we have : .iBAC=^BCA. Prop. 5. Also i. DAC= L DC A. Prop. 5. .•. z.DAB=.^DCB. Ax, 2. Similarly by joining DO it may be shown that the ^ABC=^ADC. J. Prove that the diat^onal of a rhombus bisects each of the an^^les through which it passes. From the figure in deduction (1) above we have : In the A's ABC and ADC j AB = AD. Def of a rhombus. \ CB = CD. Def. ot a rhombus. (./.ABC=^ADC. Troved in deduction 2. .'. L BAG = L. D AC. Prop. 4. And L BCA= l DC A. Prop. 4, That is, the diagonal AC bisects the angles of the rhombus through which it passes. 4. Tiik) isosceles triangles, A DC and DUC, have the same base RC. {a) Prone that the angle ABD is equal to the angle A CD. (b) Prove th it thi angle BAD is equal to ihc angle CAD. (f) Prove that AD, or AD produced, bisects the base BC. •ms>* Let ABC and DBC be the two isosceles A's en the same base BC, and let AD be joined and pro- duced to meet BC in E. It is required : (a) To prove Z.ABD: z-ABC-^ACB. ^DBC=_DCB. ,'. -lABD=^ACD. (*) To prove ^BAD = = z.ACD. -. Prop. 5. ' Prop. 5. Ax. 3. _CAD. In the A's BAD and CAD AB = AC. Def. of isosceles A' DB=DC. Def. of isosceles A- L ABD= -i. ACD. Proved above. .•. ^BAD=iLCAD. Prop. 4. (c) To prove AD produced bisects BC. In the A's ABE and ACE. AB = AC. Def. of isosceles A- AE=AE. Common. z. B A E - I. CAE. Proved above. base BE -base CE. Prop 4 { {, ■5. ABC '■" an isosceles triangle; and in the base BC two points D and E are taken such that BD— CE; prove that A DE is an isosceles triangle. B D €0 Let ABC be the given A) let the points D and E be taken such that BD = C£ and let AD and AE be joined. It is required to prove APE is an isosceles A. In the A's ABD and ACE A B = AC. Def. of isosceles A- BD = CB, Hyp. ^ ABD ^^ ACE. Prop. 5. .•.AD=AE. Prop. 4. That is, ADE is an isosceles A- 6. Prove that the diagonals of a square divide, the figure into four isosceles triangles. A ' B {. Let ABCi.) t>e the given squaie, and AC and BD the diagonals intersecting in E. It is required to prove that AEB, CEBj ADE, and CDP', are isosceles A's In the A's DAB and ABC A D = BC. Def. of a square. AB = AB. Common. ^DAB=i.ABC. Ax. 11. .•.AC = DB. Prop. 4. Again, in the A's ABE and CBE f AB = BC. Def. of a square. \ BE = BE. Common. tz.ABE= i-CBE. Proved, deduction l. .•. AE = EC. Prop. 4. Similarly we may show DE=EB. • .-. AE=EC = DE = E3. Ax. 7. That is, AEB, CEB, ADE, and CDE are isos- ce'es A's. * 7. Two equal circles, whose centres are A and B, intersect at the point C. Join CA and CB. and produce them to meet the circumferences at D and E respectively. Join DE, Prove that the an^le CDE equals the angle CED. C f 1^ 6 EUCLID I^t ACG and BCF be the two equal circles, in- tersecting at the point C. Let CA and CB be joined and produced to D and E. Let D£ be joined. It is required to prove ^CDE=lCED. CD is a diameter of the circle BCF. CE is a diameter of the circle ACG. And since the circles are equal, CD must = CE. .', DC£ is an isosceles A. Def. of an isosceles A- And :.CDE=^CED. Prop. 5. jCD be the given rhombus, and let the diagonals AC and DB intersect at E. It is required to prove AE = CE, DE = BE, and the ^'s AEB, AED, CEB, CED all right angles. IntheA's AEB, AED AL = AD. AE = AE. ^DAE=^BAE. DE = BE. Also z. AEB = /.AED. That is, each of them is s rt. L . Similarly it may be shown by taking the A'« ABE and CBE that AE = CE, and that the ^'s BEA and BEC are rt. l 's. {, Def. of a rhombus. Common. 3. Prop- 5- Prop. 4. Prop. 4. Def. 7. Ded. J. Show that the straight lines which bisect the nniiles at the base of an isosceles triangle form, with the base, a triangle 7vhich is also isosceles. Let ABC be the given isosceles A, and let the lines AD and CD bisect the angles BAC and BCA respectively. It is required to prove that ADC is an isosceles A. ^BAC=£BCA. /.DAC = halfofBAC. A.DCA=halfofBCA. .-. /.DAC= ^UCA. .-. AD = CD. That is, the A ADC is isosceles. .# * The But And Prop. c. Hyp. Hyp. Ax. 7. Prop. 6. A tf ini fei du in an I AE poi muf w DEDUCTION 41. In the figure of Prop. /, if the straight line A/i be produced both way;, to meet th: one cir- Kutnferem e at D and the other at E, show that the triangle CDE is t.wsceUs. Let ACD and BC2 be the two circles intersect- ing at C, and let AB produced meet »he circum- ference of the circle ACD in D, and let BA pro- duced meet the circumference of e c/rcle BCE in E, and let CE and CD be joined. It is required to prove that CE = CD. BC = AC. Del. of an equilateral A- :.L CAB = ^CBA. Pr 5. Also AE=AB. Def. of a tucle. And nD = AB. Def. of a circle, .'. ^E=BD. Ax. I. And AB = AB, Common. .-. E3 = DA. Ax. 2. In the A's EBC and DAC f EB = DA. Pro-.-jd. BC=AC. Proved. * I ^EBC=i.DAC. Proved. .•.CE = CD. Prop. 4. That is, CD E is an isosceles A- PROPOSITION vni. /. The opposite sides of a quadrilateral A'BCD are equal. Prove that : (a) The opposite angles are equal. (d) The angle A BD is equal to the angle CDB. {c) The middle point of BD is equidistant from A and C, Let ABCD be the given quadrilateral, having AB = DC and AD = BC, and let E be the middle point of DB. It is required to prove : (a) The z. ABC = 1. ADC and z.DAB=z.DCB. (b) The ^ABD-^CDB. (c) That AE = CE. In the A's ABD and CDB AB = DC. AD = BC. £.DB = DB. •.^DAB=z.DCB. And^ABD=^CDB. { (*) .•.^ABC=z.ADC. That is, the oppciite l 's are equal, (a). Hyp. Hyp. Common. Prop. 8. Prop. 8. XTGp. u. Ax. r;. Hyp. Again, in A's ABE and CDE AH = CD. BE-DE. .ABE=^CDE .AE-CE. (c). a. Show that two circumferences can iut eack other in only one point on the same si. '.' of the line joining their cenlrt:s. C i lor-. .. If possible, let the two circumferences CDE and CDF cut each other in two points C and D above the straight line AB which joins their centres. Join AC, AD and BC, BD. Post. 1. Then AC=AD. Def. of a c»rcle. AilBC = BD. Def. of a circle. Which is impossible. Prop. 7. Therefore two circumferences cannot cut each other in more points than one above the Ime join- ing their centres. J. Two isosceles triangles are on the same base, and on opposite sides of the base. Prove that the line joining their vertices bisects each of the verti- cal angles. c Let ACB and ADB be the two isosceles A's on the opposite sides of the base AB, and let CD be joined. It is required to prove that ZD bisects the Z.ACB and the /i ADB. In the A's ACD and BCD rAC = BC. Hyp. jAD-DB. Hyp. lCD = CD. Common. .*. L ACD = L BCD. Prop. 8. And /.ADC=^BDC Prop. 8. That IS, the z.'s ACB and ADB are bisected, 4. ABC is an isosceles triangle, of which AB and AC are equal sides. Points- D and E are taken in AB, and points F and G in AC, such tha: AD=-AF, andAE=AC. CD and BF intersect in H; CE and BG intersect in K. Prove that r (a) AH bisects lDAF. \p) i-BuII- lCI'II. (S) EH=HG. {d) A,H and K art in the same straight line. EUCLID Let ABC be the given isosceles A. having AK = AC, and le, AD = AF and AE = AG, and let CD and BF be joined, intersecting in H. («) AH bisfcts the It is required to prove that /.DAF. In the A's BAF and CAD AB-AC. Hyp. AF = AD. Hyp, L BAF = L CAD. Common. .•.BF = CD. Prop. 4. And ^ABF=^ACD. Prop. 4. But iLABC^^ACB. Prop. 5. /. ^HBC=^HCB. Ax. 3. .-. HB= HC. Prop. 6. .*. FH=. DH. Ax. 3. Then in the A's AD H and AFH AD = AF. Hyp. !AH=AH. Common. [DH = FH. Proved. ^DAH = ;lFAH. Prop. 8. (*) It is required to pro"e /LBDH=iLCFH AB = AC. AD=AF. /.DB = FC. Then in the A's DBH and FCH DB = FC. DH = FH. HB = HC. .•.^BDH=^CFH. It is required to prove EH = HG. Join EH and GH. AE = AG. AD = AF. .•.DE = FG. Then in the A's DEH and FGH { DE = FG. \ DH=FH Uedh=.lgfh. .•.EH = HG. Hyp. Hyp. Ax. 3. Proved. Proved. Proved. Prop. 8. Post. I. Hyp. Hyp. Ax. 3. Ax. 3. Proved. Proved. Prop. 4. kd) It is required to prove A, H and K are in the same straight: line. In theA'sBAKandCAK AB = AC. .•.BK=CK, AK = z.BAK = = AK. ^CAK. Hyp. Common. Proved. Prop. 4. Tnerefore in A's BHK and CHK i BH = CH. \ HK = HK. ( BK = CK. .'. ^BHK=.lCHK. And in A's BDH andCFH BD = CF. DH = FH. BH = CH. ,*. z.DHB=i.FHC. Again, in A'sAHD and AHF AD=AF. AH=AH. DH=FH. .•. A.AHD=^AHF. .-. ^'s AHD, DHB and BHK= FHCandCHK. That is, ^'s AHD, DHB, and BH four right angles, or two right angles, a straight line. Proved. Common. Proved. Prop. 8. Proved. Proved. Proved. Prop. 8. Hyp. Common. Proved. Prop. 8. = z.'s AHF, Ax. 2. K are half of and AH K is PROPOSITIONS IX. AND X. /. To divide a given an^le into four equal parts. B Let ABC be the given L. It is required to divide it into four equal parts. Bisect I. ABC by BD. Prop. 9. " ^ABDby BE. Prop. 9. " ^DBCbyBF. Prop. 9. Then ^'s ABE, EBD, DBF, .tnd FBC are all equal. Ax. 7. .'.the .:.ABC has been divided -nto four equal parts. 2. To divide a given straight line into eie;ht equal parts. A H £ I B Let AB be the given straight line. It is required to divide it into eight equal parts. Bisect AB at C. Prop. 10, " AC " D. Prop. 10. " CB " E. Prop. 10. " AD " F. Prop. 10. " DC " G. Prop. 10. " CE " H. Prop. iu. «« EB " I. Prop. 10. Then AF, FD, DG, GC, CH, HE, EI and IB are all equal. Ax. 7. Then the straight Sine AB has been divided into eight equal paits. DEDUCTION J. On a given base describe an isosceles triangle, such that the sum of its equal sides shall be equal to a given straight line. C Bisect AB ut D. Prop. lo. Then DB is >4 of AB. Bisect DBat F. Prop. lo. Then FBis H of AB. Bisect AD at B, then AE is % of AB. With centre B and distance BF describe a circle FGC. Post, 3. Produce AB to meet the circumference in C. Post. 2. AC is the line required. AB is divided inio four equal parts : AE, ED, DF, and FB. But BC = FB. .". BC is = io ^ of AB. .". BCis^ofAC. But BC is the part produced. .'. AC is the line required. Let AB be the given base, and CD the given straight line. It is required to describe on AB an iiosceles A, the sum of whose sides shall = CD. Bisect CD at F. Prop. 10. ThenCF = FD. Ax. 7. From A draw AG = CF. Prop. 2. " B " BH = FD. Prop. 2. With centre A and radius AG describe the circle GEO. Post 3. With centre B and radius BH describe the circle HEO. Post. 3. Let the circles cut in the point E. Join EA and EB. Post. i. EAB is the A required. GA=CF. GA=AE. .•.AE = CF. BH = FD. BH = BE. .•.BE = FD. But CF and FD = CD. AndCF " FD = GAandBH, .'.GAand BH = CD. But GA and BH = EA and EB .'.EAand EB = CD. ButCF = FD. .".GA and BH are equal. And AE and EH are equal. Ax. I. .'.AEB is isosceles. Def. of isosceles A- And AE and BE are equal to CD. Proved. 4. Produce a giiien straight line so that the whole line may be five times as long as the part produced. J. D, E and F are the middle points ./the sides of an equilateral triangles show that the triangle DEF is equilateral. Const. Def. of a circle. Ax. I. Const. Def. of a circle. Ax. I. Const. Ax. I. Const. Ax. I. Ax. 7. Let ABC be the equilateral triangle and D,E,F the middle points in the sides AB, BC, CA. And let DE, EF, and FD be joined. It is required to prove DEF equilateral. In the A's DAF and FCE AF = CF. Ax. 7. A!) = CE. Ax. 7. .^DAF=^ECF. ^'sofequil.A .-. DF = EF. 1.4. In the A's DDE and CEF BD = CF. Ax. 7. BE = CE. A.x. 7. ,L DBE = _ FGE. L 's of equil. A DE = EF. 1.4. DE, EF and FD are all equal. Ax. 1. DEF is an equilateral triangle. {. Let AB be the given straight line. It is required to produce it so that the part pro- duced shall be one-fifth of the whole line thus pro- duced. 6. Two isosceles triangles ARC and DEC stand on the same base BC but on opposite sides of it. /■: is the middle point of AB, and F the middle point of A C, and BF and CE intersect tn G. (a) Prove that DE = DF. (b) Prove that ^EDH=^FDC. (c) Prove that CE = BF. (d) Prove that z.DEF=/.DCE. (e) Prove that AGBC is isosceles. (/) Prove that EG = GF. ig) Prove that ^ EG D = i. FGD. 10 EUCLID 'Jt) Prove that A, G straight line. and D are in the same Post. I. Prop. 5. Prop. 5. Ax. 2. Let ABC and DBC be two isosceles A's on the same base EC but on opposite sides of it, having E the middle point of AB and F the middle point of AC. Let BF and CE be joined and mtersect at G. («) It is required to prove DE = DF, Join DE and DF. ^ABC=z.ACB. ^DBC=^DCB. .•. ;lABD=i.ACD. In A's t>BEandDCF D B = DC. Def. of an isos. A. BE = CF. Ax. 7. .EBD=z.FCD. Proved. .M)E = DF. Prop. 4. It is required to prove z.EDB==- _FDC. In As DBE and DCF f DB = DC. Def. of an isos. A- \ BE = CF. Ax. 7. (.z.EBD=^FCD. Proved. .'. ^EDB=_FDC. Prop. 4. It is required to prove CE = BF. InAsBAFandCAE BA = CA. Def. of an isos. A. AF = AE. Ax. 7. L BAF = L. CAF. Common. CE=BF. Prop. 4. {d) It is required to prove l DBF= l DCE. In A's DBF and DCE fDB = DC. ■^BF = CE. Idf=de. :. ^dbf=^dce. {, Def. of an isos.A- Proved in {c). Proved in {a). Prop. 8. (0 isosceles. In A'8 BFC and CEB fBF = CE. Proved in (^ of the four angles at the point G. Or i_'s EGA and EGD = 2 straight angles. That is, A, G and D are in the same straight line. PROPOSITION xi. /. In what line do all joints lie which are equally distant from a given point f Let A be the piven point and B, C, and D three points equally distant from A. It is required to find th: line in which all points would lie which are at the same distance from A as B, C, and D. Therefore AB, AC, and AD must be radii of the same circle whose centre is A. Def. of a circle. Therefore all points which are at the same dis- tance as B, C and D muiit be on the circumfer- ence of the circle BCD. DEDUCTION II 2. Find a line in which all points lie which are 4. Give a construction for finding the centre of a equidistant from two given points. circle whichpasses through the three angular points of a triangle. Let A and B be the two given points. It is required to find a line, all points in which are equally distant from A and B. Join AB. Bisect AB in C. Draw CD ± to AB. Then CD is the required line. In CD take any point K. Join AK and BK. Then in the A's ACK and BCK AC = BC. CK = CK. ^ACK=aBCK .•.AK = BK. Similarly it may be shov.-= that any other point in CD is equally distant from L and B. Post. I. Prop. 10. Prop. II. Post I. Construction. Common Ax. I . Prop, 4. 3. Find, 1/ possible, a point which is equidistant from three given points. Let A, B and C be the three given points. It is required to find a point equidistant from A, B and C. Join AB, BC, AC Bisect AB in E. Bisect BC in G. Erect EO ± to AR. Etect GO J. to BC. O shall be the required point. Join AO, BO, CO. In the A's AEO and BEO AE=BE. EO = EO. ^OEA=-^OEB. .'.A0 = OB. Again, in the A's BOG and COG CG = BG. Const. n.f\ —.nf\ r^ ^OGC=^OGB Def. ofa±. .'.OBr^OC Prop. 4. .•.OA=OC. Ax. I. That is, O is equidistant for A, B and C. AS {, / I Post. I. Prop. 10. Prop. 10. Prop. II. Prop. II. Post. I. Const. Common . Def. J.. Prop. 4. Let and C be the three angular points of A^BC. It is required to give a construction to find the centre of the circle which will pass through A, B and C. Bisect AB in E and AC in F. Prop. i. Erect EO ± to AB and FO to AC. Prop. n. O is the centre required. Join AO, BO and CO. Post. i. In A's AEO and BEO AE = BE. Const. EO = EO. Common. Z-AEO=£.BEO. Def. of a J.. .•.AO = BO. Prop. 4. Again, in the A's AFO and CFO r AF = CF. Const. \ FO = FO. Common. U AFO =^ CFO. Def. of a ±. .•.AO = OC. Prop. 4. .•.BO = OC. Ax. I. With centre O and distance OB describe a circle ABC. Post. 3. The circumference will pass through A, B, C. Def. of a circle. 5. Find a point whose distance from a given point A is equal to a given straight line, and. whose distance from a given point B is equal to another straight line. : V- ana u oe tne two given straignt imes. is required to find a point whose distance a giren point A is equal to a given straight line D, and whose distance from a given point B is equal to another straight line C. It from 13 EUCLID From A draw AE = D. Prop. 2. From B draw BF = C. Prop. 2 With centre A and distance AE describe a circle EGH. Post. 3. With centre B and distance BF describe a circle FGH. Post. 3. Let the circles intersect at G and H. loin AG and BG. Post. i. Then G shall be the required point. Because A is centre of circle EGH AE = AG. Def. ofci.cle. But AE = D. Const. .•.AG = D. Ax. I. Because B is centre of circle FGH BF = BG. Def. of circle. But BF = C. Const. .•.EG = C. Ax. I. This is not possible when the distance between the two points is greater than the sum of the given straight lines. 2. In a given straight line find a point that is equally distant from two given points. PROPOSITION XII. /. Describe an isosceles triangle, having given the base and the length of the perpendicular drawn from the vertex to the base. Let A be the given base and B the given per- pendicular. It is required to describe an isosceles triangle having its base equal to A and perpendicular equal to B. Take any straight line CD limited at C un- limited towards D. From CD cut of!" CF=A. Prop. 3. Bisect CF in E. Prop. 10. From E erect a perpendicular EG. Prop. 11. From EG cut off E H = B. Prop. 3. Join CH and FH. Post. i. Then shall HCF be the required triangle. IntheA'sHCEandHFE CE = FE. Const. EH = EH. Common. .HEC = HEF. Ax. 11. .•.CH = Fh. Therefore HCF is an isosceles A having its base CF = to the given straight line A, and its perpendicular height EH= to the given line B. Def. of isos A- I Let AB be the given straight line and C and D the two given points. It is required to find a point in AB that will be equally distant from C and D. Join CD. Post I. Bisect CD in F. Prop. 10. From F draw a perpendicular to CD, which meets AB in G. Prop. 11. Then G shall be the point required. Join CG and DG. Post. i. In A's CFG and DFG CF = DF. Const. FG = FG. Common. ^CFG=z.DFG. Ax. 11. .•.CG = DG. Prop. 4. Therefore a point G has been found in the straight line AB, which is equally distant from C and D. Note ; When the given points are on opposite sides of the given straight line and not eq^ually distant from it, and when the straight line joining them cuts the given straight line at right angles, the proposition is impossible. 3. From two given points on opposite sides of a given straight line, draw straight lines to a point in the given line, making equal angles with it. Let AB be the given straight line and C and D the given points on opposite sides of AB. It is required to draw from C and D straight lines to a point in AB, making equal angles with it. From the point D draw DE perpendicular to AB. Prop. 12. Produce DE to F. From EFcut EG = ED Join CG. Produce CG to meet AB in H. Then H shall be the required point, loin DH. Then in A's GEH and UEH f EG = ED. \ EH = EH. Ugeh=z.deh. .-. :-GHE=iLDHE. Post. 2. Prop. 3. Post. I. Post. 2. Post. I. Const. Common. Ax. II. DEDUCTION »3 Therefore a point H has been found in the straight line AB, such that CH and DH make = z. 's with AB at that point. Prop. 4 4, ABC is a triangle and D, E and F are the middle points of the sides BC, CA, and AB re- spectively. From D a straight line is drawn per- pendicular to BC, and from E another straight line is drawn perpendicular to CA, meeting the former line in O. Show that OF is perpendicular toAB. Let ABC be the given t;iangle, and D, E and F the middle points in BC, CA and AB respect- ively, and let a line be drawn from D perpendicu- lar to BC, and from £ a line perpendicular to CA, and let these Imes meet in O. . It is required to prove that OF is perpendicular to AB. Join AO, BO, CO. Then in As ODB, ODC BD = CD. DO = DO. .LOTm= lODC. BO = CO. In A'sOCEandOAE. AE=CE. OE=OE. ^ of a rt. angle. .•.^.ABD+^ADB = i rt, anrle. And u DAB + ^'s ABD and ADB = 2 rt. angles. 4. Construct an angle equal to half a rt. angle. Take a straight lin; LC. From B draw a perpendicular BA. Bisect L ABC by the Jine BD. L ABD is the required angle. ^ ABC is a rt. angle. And aABD=z.DBC. Therefore iLABD = ;^ofart. angle. Prop. II. Prop. 9. Def. I ; Const. 5. Make an isoscles triangle having each of its base angles equal to half a right angle, and each of the equal sides equal to a given straight line. Let A be the given straight line. It is required to make an isosceles triangle hav- ing each of its base angles equal to half a right angle, and each of the equal sides equal to A. Take any straight line BC terminated at B, but unlimited toward C. From B draw a ± BD to BC. Prop. 1 1. From BC cut off BE = A. Prop. 3 From BU cut off BF=A. Proa ^ Join FE. Then EBF shall be the triangle required. In A EBF /'z.FBE=i rt. L, i.BFE=Jrt. L. Ded ^BEF=irt. L. Ded. BF=A. . BE=A. Therefore FBE is an isosceles triangle, having each of its base u 's equal to Yz rt. l and each of its equal sides equal to the straight line A. Def. 10. 3, above. 3, above. Const. Const. 6. If one of the four angles which two intersecting straight lines make with each other be a right tingle, all the other angles are ri^ht angles. A B D Let AB and CD be two intersecting straight lines which cut at E, making the l. AEC = i rt angle. It is required to prove L CFB = i rt. angle, L BED = i rt. angle L DEA=i rt. angle. L AEC + L BEC = 2 rt. angles. But Z.AEC = i rt. angle. .'. Z.CEB = i rt. angle. Z.CEB+ £.BED=2 rt. angles. But -i.CEB=i rt. angle. Prop. 13. Hyp. Ax. 3. Prop. 13. Proved. Ax. 3. angles. Prop. 13. Proved. -_™„.^.,™.„,,„ ,- ,^ .„„^ Ax. 3. That ll, ^CTB=i rt. angle, BED=I rt. angle, ^ of 4 rt. angles = 2 rt. angles. That is GE and HE are in the same straight line. I- 14. 4. Show that if AB is perpendicular to the straight line CD, which tt meets at B, then if AB is produced to £■, BE is also perpendicular to CD. A Let AB be a straight line X'to CD and meeting CD in B. It is required to prove that if AB be produced to E, that EB is J, CD. z.ABC==^ABD. Def. of±. Then ^ADC. And ^BCD> iLBAD. Join AC. Post. 1. IiiAABC, iL BCA is greater than ^.BAC. Prop. 18. In ADAC, ^DCA is greater than ^DAC. Prop. 18. Therefore ^BCD is greater than ^DAB. Ax. 4. Similarly, by joining BD it may be shown that z. ABC is greater than /.ADC. PROPOSITION XIX. I. In an obtuse-angled triangle the greatest side is opposite the obtuse angle ; and in a right-angled triangle the greatest sid is opposite the right angle. (a) Let ABC be the obtuse-angled triangle, and let £,ABC be the obtuse angle. It is required to provs that AC is greater than either AB or BC. la AABC the ^'sA + B are less than 2 rt. angles. ^ Prop. 17. But z. B is an obtuse angle. Hyp. .'. L A must be less than a rt. angle. Similarly we may show the lQ lets tnan a rt. angle. That is, L H is greater than either ^A or ^C. .*,AC is greater than AB or BC. Prop. 19. {b) Let DEF be the right-angled triangle hav- ing angle DEF a rt. angle. It is required to prove that UF is greater than DE or EF. InADEFthe ^'sD-|-E are less than two rt. angles. Prop. 17. But z. E is a rt. angle. Hyp. .'. L D must be less than a rt. angle. Similarly we may show tha'. /. F is less than a rt. angle. That is, ^ E is greater than ^. D or ^ F. .'.DF is greater than DE or EF. Prop. 19. 3. Show that three equal straight lines cannot be drawn from a given point to a given straight line. Let AB be the given straight line and C the given point, and let CD and CE be two equal straight lines drawn from C to AB, and if it be possible let CF be drawn from C to AB equal to CD or CE. It is required to prove CF is not equal tu CD or CE. Inthe ADCE because CD = CEthe aCDE = /.CED. Prop. 5. Inthe ACDF because CD = CF the lCDF = ^CFD. Prop. 5. Therefore the ^CFD=iLCED. Ax. i. But ^CFD is greater than ^CED. Prop. 16. That is, the z. CFD is both equal to and greater than ^CED. Which is impossible. Therefore CF is not equal to CD or CE. Similarly it may be shown that no other straight line can be drawn from C to AB equal to CD or CE. 3. The perpendicular is the shortest straight line that can be drawn from a given point to a given straight line; and of others, that which is nearer to the perpendicular is less than the more remote. G C DEUUCriON «9 From the given point A let there be drawn to the given straight line HC (i) the perpei 'icular AD, (2) AE and AF squally distant from the per- pendicular, that is, so that DE-DF, (3) AG more remote than AE or AF. It is requiked to prove AD the least of these straight lines, and AG gieater than AE or AF. In the triangles ADE and ADF fAD = AU. Common. ^DK=DF. Hyp. UaDE=<_ADF. Ax. k .•.AE = AF. Prop, 4. Because lADE is right .'. AED is acute. Prop. .".AE is greater than AD. Prop. Hence also AF is greater than AD. Because lAJLG is greater than ^ ADE. Prop. .■.AEG is obtuse. .".AGE is acute. Prop. .'.AG is greater than AE. Prop. Hence also AG is greater than AF, nnd than AD. 17- «9 16. 17. 19- Enunciation. — Any straight Ime drawn from the vertex of an isoscbles triangle to a point in the base produced is greater than either of the equal sides. Let ABC be the given isosceles triangle, having BC produced to F, and let AF be joined. It is required to prove that AF is greater than AB or AC. In AABC the -lABC= A.ACB. Prop. 5. The lACF is greater than ^ ABC. Prop. 16. Therefore ^i ACF is greater than lACB. But ^ACB isgre.uerthan i.AFC. Proo. i6. Therefore ^ ACF is much greater tha-i ^AFC. Then AF is greater than AC or AB. Prop. 19. 6. The vertical angle ABC of the triangle ABC is bisected by the straight line BD, which meets the base in D. Show that AB is greater than AD, and CB is greater than CD. 4. Any straight line drawn from the vertex of an isosceles triangle to a point in the base is less than either of the equal sides. D C Let ABC be an isosceles triangle, and let AD be the straight line drawn from the vertex to the base. It is required to prove AD is less than AB or AC. The z.ADC is greater than the ^ABD. Prop. 16. But ^ABD=^ACD. Prop. 5. Therefore z. ADC is greater than L.ACD. That IS, AC is greater than AP. Prop. 19. Similarly it may be shown that AB is greater than AD. 5. Enunciate, and prove a theorem similar to Ex. 4 when the point is taken in the base produced. Let ABC be the given triangle, having L ABC bisected by BD, which meets AC in D. It is re- quired to prove that AB is greater than AD, and that CB is greater than CD. The ^ABD=lCBD. And L BDA is greater than l CBD. Then z. BDA is greater than ^ABD. Therefore AB is greater than AD. Similarly it may be shown that CB than CD. Consi. Prop. 16. Prop. 19. is greater PROPOSITION XX. r Prove Prop. 20 by bisecting the vertical angle V a straight line which meets the base. be the triangle and let l. BAC be the straight line AD, which meets Let ABC bisected by BC at D. it is required to prove that BA-fAC is greater than BC. The z. ADD is greater than ;lCAD. Prop. 16. Bur/.CAD= IbAD. Hyp. /. lADB is greater than Z-BAD. .•.AB is greater than BD. Prop. 19. i. ADC is greater than l DAB. Prop. 16. I) •e EUCLID But iLDAB=/.DAC. Hyp. .'. :. ADC is Rr«ater than L DAC. .", Therefore AC is Hreater than CD. Prop. 19. .•.AB + AC is greater than BD + DC=»BC. Ax. 4- *. Prove Prop. 20 by drawing a perftndicular from the vertex to the base. 4, (.1) Prove that the sum of the sides of any quadritattral is neater than twice either diagonal Let ABC be the triangle, having AD the per- pendicular drawn from A to the base BC. It is required to prove that BA+AC is greater than BC. Z.ADB is a right angle. .'. L DAB is an acute angle. .'. ^ ADB is greater than the i^DAB. .".BA is greater than BD. - ^ADC is a right angle. .'. L DAC is an acute angle. .'. ^ ADC is greater than the l DAC. ,*.AC is greater than DC. ,. .•.AB+AC is greater than BD + CD = BC. Ax. 4. J. (a) In the figure of Prop. 16 prove that CF is equal to AB. A >^R Hyp. Prop. 17. Prop. 19. Hyp. Prop. 17. Prop. 19. Let EB = EF and AE = EC. It is required to prove that CF=AB. In the A's ABEand FCE fAE = EC. Hyp. ■^BE = EF. Hyp. UaEB=^FEC. Prop, li .•.CF = AB, Prop. 4. J. {b) Hence prove that the sum of any two sides of a triangle is greater than twice the straight line drawn from the middle taint of the third side to the opposite vertex. Use the same figure m 3 (a). Le' >ir = AB. Proved in (a)^ i. .. requirrid to prove AB + BC is greater than twice BE^ Bc + uf is gieaifcr than lii'. Prop. 20. But BC + CF = BC + AB, since CF = AB. And BF = twice. BE, since BE = EF. That is, AB + BC is greater than twice BE. Let ABCD be the given quadrilateral and DB and AC its diagonals. It is required to prove that AB + BC + CD + DA is greater than twice AC or twice DB. In the AADB the sides AD + AB are greater than DB. Pfop. 20. IntheACDBthe sIHcs CB + CD are jjreaier than DB. Prop. 20. That is, AB + AD + BC + DC is greater than twice DB. Similarly it may be ohown that AB + BC + CD + DA is greater than twice AC. 4. (b) Hence prove that the sum of the sides is greater than the sum of diagonals. Use same iiKure as 4 (a). The sides AB + BC + CD + DA are greater than twice DB. Proved. The sides AB + BC + CD + DA are greater than twice AC. Proved. That is, twice (AB + BC+CD + DA) is greater than twice (DB + AC). Therefore AB + BC + CD + DA is greater than DB + AB. That is, the sum of the sides of the quadrilateral is greater than the sum of the diagonals. J. Take any point O. and join to the angular points of the triangle ABC. (a) Prove that the sum of OA and OB is theater than AB. (b) Prove that twice the sum of OA, OB and OC Tf greater than the sum of the sides. Let O be the given point and ABC the given trian"!* of which the angular "oints A; B and C are joined to O. It is required to prove that the sum of OA and OB is greater than AB. M DEDUCTION 31 4 AO and OB are any two tides of the triangle AOB. Therefore AO + OB is greater than AB. Prop 30. (i) Use same figure as in (a). Let O be the given point and ABC the given triangle, of which the angular points A, B and C are joined to O. It is required to prove that twice the sum of OA, OB and OC is greater than the sum of AB, AC. BC. kO + OB is greater than AB. Prop. 20. O- '"' is greater than BC. Prop. 20. OA + OL is greater than AC. Prop. 20. That is, twice OA + OB + OC is greater than once AB + BC + CA. 6. If a point be taken within a quadrilateral and joined to each of the angular points ^ show that the sum of these joining lines is the least possible, when the point taken is the point of intersection of the diagonals. Let ABCD be the quadrilateral, and E any point in it, and F the point of intersection of the diagonals. It IS required to prove that AE, BE, CE and DE are greater than DB and CA. DE + EB is greater than DB. Prop, 2a AE + EC is greater than AC. Prop. 20. Therefore AE + BE + CE + DE is greater than AC and DB. 7. A f>oint P is taken within the triangle ABC. Show that the sum of the sides AB and AC is greater than the sum of PB and PC, Let ABC be the given triang1»1»d P the given point within it, and let BP and CP be joined. It is required to prove B.\ + AC is greater than BP + PC. Produce BP to meet AC in E. Post 2. Then BA + AE is greater than BE. Prop. 20. Add EC to each. Then BA + AC is greater than BE + EC. Ax. 4. Again, PE + EC is ^reaterthan PC. Prop. 20, Add PB to each. Then BE + EC is greater than BP + PC. Ax. 4. But BA + AC is greater than BE + EC. Proved. Much more then is BA+AC greater than BP + CP. S. Four points lie in i plane, no one of them being within the triangle formed by joining the other three. Find the point thu sum of wfiose dis- tances from these four points is the least possible. Since no one of t^u points is within the tri- angle, the four points when joined form a quad- rilateral. Let ADBC be the quadrilateral. It is required to find a point the sum of whose distances from A, B, C, D is the least possible. Join AB and CD. Post. i. Let E be the point where they intersect. Then E is the point required. Deduction 6. g. In any triangle, the difference between any two sides is less than the third side. Let ABC be the given triangle. It is required to prove B.\ greater than the iArA»>Am/*A K«tt»***% T^^ ^nd J^kCZ BA+AC is greater than BC. Prop. 20. Take AC from each side. Then BA is greater than BC - AC Ax. 5. 22 EUCLID PROPOSITION XXI. /. In the figure of Prop, si, join DA, and show that the sum of DA, BD and DC is less than the sum of the sides of the triangle ABC^ Intt greater than half the sum. Let ABC be the triangle, and from B and C let BD and CD be drawn to any point D within the triangle, and let AD be joined. It is required to prove (i) DA + BD + DCless than BA+AC + CB. (2) DA + DB + DC greater than BA + AC + CB. 2 (x) AD + DB isless than AC + BC. Prop. 2r. BD + DC is less than B A + AC. Prop. 2 1 . AD + DC is less than BA + BC. Prop. 21. That is, 2(AD + BD + DC) is less than 2 (BA + AC + BC),orAD + BD + DCislessthan BA+AC -■ BC. (2', AD + DB is greater than AB. Prop. 20. BD + DC is greater than BC. Prop. 20. DC + AD is greate: than AC. Prop. 20. That is, 2 (AD + DC + DB) is greater than AB + BC + AC, or AD + DC + BC is greater than AB + BC + AC. 2. In the figure of Proposition 21, show that the angle BDC is greater than the angle BAC, by joining AD and producins[ it towards the base. Let ABC bL a triangle, and fiom B and C let BD and CD be drawn to any point D within the tri- .._„l^ ^^A !..> A Tt v.. ir^,-r.^A r.^A ^,^A,.^m.A »^ ~.»» the base BC in E. It is required to prove that the angle BDC is greater than the angle BAC. The ^BDE is greater than the lBAD. Prop. 16. The ^ CDE is greater than the z,CAD. Prop. 16. Therefore the whole u BDC is greater than the whcle L BAC. Ax. 4. PROPOSITION XXIII. /. Prove proposition 23, gi'"ing all the construc- tion instead of assuming proposition 22. Let AB be the given straight line, A the given point in it, and z. DCE the given angle. It is required to make at A an angle equal to I. DCE. In CD, CE take any points D, E. Join DE. Post i. From AB cut off AF = CD, AG = CE, and from FB cut off FH = DE. Prop. 3. With centre A and distance AG, describe circle GKL. Post. ,3. With centre F and distance FH, describe circle HKM. • Post 3. Let the circles cut each other at K. Join KF, AK. Post. I. Then AFK is the required triangle. AF = CD. Const. Because AG=:AK. Def. 1 AndAG = CE. Const. .•.AK = CE. Ax. I. Because FH = FK. Def. II '. AndFH = DE. Const. .•,FK = DE. Ax. I. .'.A AFK has its sides respectively equal to CD, CE and DE. fCD=AF Const. In the A's CDE, AFK ■ CE = AK Proved. DE=FK Proved. .•.^FAK=/.DCE. Prop. 8. 2. Construct a triangle, having ^ven the two sides and the angle between them. Let A and B be the two given sides and C the given angle. DEDUCTION »3 * < It is required to construct a triangle having its two sides equal to A and B, and the contained angle equal to C. Take any straight line DE=A. At the point D make the angle EDF» ^C. I. 23. From DF cut off DG = B. Join EG. Then EDG is the required triangle, fDE = A In the triangle EDG ]dG = B (.^EDG=Z.C Therefore the triangle EDG has its two sides respectively equal to the two given sides and the contained angle equal to the given angle. J?. Construct a triangle, having given the base, and having the angles adjacent to the base equal to two given angles. Is this always possible f A 2. Assuming the truth of proposition 23 J deduce the truth of proposition 24. 1.3. Post. I. Const. Const. Const. Zb Zl 23- 23. Let A be the given base and B one angle at the base and C the other angle at the base. It is required to construct a triangle, having the base equal to A and the angles at the base equal to B and C. Take any straight line EF = A. At E make an angle equal to B. At F make an angle equal to angle C. Produce EG and FG, meet at G. Then EFG shall be the triangle required. The angle at E = angle B. The angle at F = angle C And these are ths angles adjacent to the base. And the base EF was made equal to the given line A. It is impossible to construct the triangle if the given angles are right angles or greater than right angles. U- '7') Const. Const. PROPOSITION XXV. /. Show that Propositions 24. and 25 are con- verse propositions. Converse propositions are those in which the conclusions of each becomes the hypothesis of the other. • L 1 In the 24th proposition we are given the angles contained by the cqoal sides of two triangles, un- equal ; and are required to prove the base of the triangle having the greater contained angle greater than the base of the other Itiangie. In the 25th proposition we are given the bases of two triangles, unequal ; and are required to prove that the angle contained by the equal sides of the triangle that has the greater base is greater than the angle contained by the equal sides of the other triangle. Let ABC and DEF be two triangles, having AB = DE; AC = DF, but ^ A greater than ^D. It is required to prove BC greater than EF. If BC is not greater than EF, it must be either equal to, or less than, EF. But BC isnot = EF. For then L A would equal L D. Prop. 8. Which it does not. Hyp. Also BC is not less than EF. For then ^ A would be less than L D. Prop. 25. Which it is not. Hyp. Therefore BC is greater than EF. 3. D is the middle point of the side BC of a tri- angle ABC. Prove that the angle A DB is greater or less than the angle ADC according as AB is greater or less than A C. CASE I. Let ABC be the given triangle, having D the middle point of BC, and let AD be joined, and let AB be less than AC. It is required to prove ^ADB is less than .lADC. In the triangles ADB and ADC, F"» = CD. AD = AD. But AB is less than AC. Then ^ ADB is less than ^ADC. Hyp. Common. Hyp. Prop. 25. CASE II. 34 EUCLID Let ABC be the given triangle, having D the middle point of BC, and let AD be joined, and let AB be greater than AC. It is required to prove Z.ADB greater than iiADC. In the triangles ADB and ADC, BD--CD. Hyp. AD=AD. Common. But AB is greater than AC. Hyp. Then z. ADB is greater than Z.ADC. Prop. 25. ,*. L ADB is greater or less than ^ ADC accord- ing as AB is greater or less than AC. 4. D is the middle point of side BC of a triangle ABC. Prove that, if uADB is greater than L.ADC, then /. 7 is greater than AC, and, if lADC is greater than lADB, then AC is try eater than AB. CASE I. Let ABC be the triangle, having D the middle f)oint of the side BC, and let AD be joined, and et L ADB be greater than i. ADC. It IS required to prove that AB is greater than \C. In A's ADB and ADC, AD = AD. Common. BD = CD. Hyp. But L ADB is greater than Z.ADC. Hyp. Then AB is greater than AC. Prop. 24. Let ABC be the triangle, having D the middle ftointof the side BC, and let AD be joined, and et /L ADC be greater than ^ ADB. It is required to prove that AC is greater than AB. In Aa AUB and AuC, AD = AD. Common. BD = CD. Hyp. But lADC is greater than Z-ADB. Hyp. Then AC is greater than AB. Prop. 24. PROPOSITION XXVI. /. The angle BA C is bisected by the straight line AD. From D the lines BD and DC are drawn making the angles ADB and ADC equal. Prove that DB is equal to DC. ai si C Let BAC be the angle and let it be bisected by AD. From D let DB and DC be drawn, making iLADB=^ADC. It is required to prove DB = DC. /■AD = AD. Common. InA'sADBand ADC-^i.DAh=^DAC. Hyp. UaDB=^ADC. Hyp. /. DB = DC. Prop. 26. s. The bisector AD of the angle BAC of the triangle ABC meets BC in D. Prove that, if the angles ADB, ADC are equal, the triangle is isosceles. Let ABC be the given triangle, having its angle BAC bisected by AD, which meets BC at D, and let z.ADB=z.ADC. It is required to prove that AB = AC. f/.BAD=^CAD. Hyp, i_ tt.. A>-Apr» Afn' ' Aptl— , Kwn u..- \ AD = AD. Common. .'. AB=AC. Prop. 26. That is, A ABC is isosceles. Def. of an isosceles triangle. t1 C DEDUCTION 25 Hyp. Hyp. rop. 26. Hyp. )mmon. J. The equal angles of an isosceles triangle ABC are bisected by BD and CE, which meet the oppo- site sides in D and E. Prove that BD is equal to CE. Let ABC be the given isosceles triangle, and let the angles ABC and ACB be bisected by BD and CE, which meets the opposite sides in D and E. It is required to prove BD = CE. The z. ABC =^ ACB. Therefore ^ABD = ACE. In »»,-fr;o«„i-cr^ABD=z.ACE. aV^VJ^^^^J't^J z.BAD= lCAE. ACEandABDl BD = CE. AB = AC. Prop. 5. Ax. 7. Proved. Common. Hyp. Prop. 26. 4. Any point in the bisector of an angle is equi- distant from the arms of the angle. Let BAC be the angle and AD the bisector and D anV point in it. It, is required to prove that D is equi-distant from AB and AC. From D drop a perpendicular DF on AB meeting AB in F. From D drop a perpendicular DE on AC meet- ing AC in E. r^FAD=_EAD. Hyp. in ^'S Ar u and :\i.lj \ _ .-ir :^- ^ l. .-liij--. n.-.. 1 1. t AD=AD. Common. Therefore DF = DE. Prop. 26. That is, the point D is equi' distant from AB and AC. 5. Find the point in the base of a triangle which is equidistant from the sides. Let ABC be the given triangle. It is required to find a point in BC equidistant from the sides AB and AC. Disect /lBAC by the straight line AD which meets BC at D. Prop. 9. Then D is the required point. From D drop the perpendiculars DF and DE which meet the sides AB and AC in F and E. Prop. II. Then in the A's AFD and AED AD = AD Common, Vi aa Rv K C. i^ T. I Rr,/->t Everything specially prepared for this volume. Contains broom drill, hoop drill and march, Mother Goose reception and drill, doll drill, new tambourine drill, etc. Humorous Speakers and Dialogues, Books for Holidays and Sun ''ay- Schools, Ta ..eaux, JVIono- logues, etc. Twenty-three Titles. Paper Binding, each, 30 Cents. Good Humor. For Readings and Recitations. By Henry Firth Wood. Many of the pieces make their first appearance in this volume, while a num- ber of others are original creations of the compiler. Choice Humor. For Readings and Recita- tions. By Charles C. Shoemaker. One of the best and most popular humorous recitation books ever published. Choice Dialect. For Readings and Recita- tions. By Charles C. Shoemaker. Contains se- lections in all dialects, such as Irish, Scotch, French, German, Negro, etc., representing all phases of sentiment, the humorous, pathetic, and dramatic. Choice Dialogues, By Mrs. J. W. Shoe- malcer. This is doubtless the best all-round dia- logue book in print, being adapted as it is to the Sunday-school or Day-school, to public and pri- vate entertainments, and to young people or adults. Humorous Dialogues and Dramas. By Charles C. Shoemaker. All the dialogues are bright and taking, and sure to prove most success- ful in their presentation. They can be given on any ordinary stage or platform, and require noth- ing difficult in the way of costume. Classic Dialogues and Dramas. By Mrs. J. W. Shoemaker. Contains scenes and dialogues selected with the greatest care from the writings of the best dramatists. It is rarely, if ever, that such a collection of articles from the truly great writers is found in one volume. Sterling Dialogues. By William M. Clark. The dialogues in this book were chosen from a large store of material, the contributions having been received from the best qualified writers in this field of literature. Model Dialogues. By William M. Clark. Every dialogue is full of life and action. The sub- jects are well chosen and practical, and so vjiried as to suit all grades of performers. Standard Dialogues. By Rev. Alexander Clark, A.M. In variety of subject, and adapta- tion to occasion, this book has special points of merit, and the dialogues will be found both inter- esting and instructive. Schoolday Dialogues. By Rev. Alexander Clark, A.M. Contains much good material for the young folks as well as for the older people, and furnishes a great variety and diversity oi senti- ment. Popular Dialogues. By Phineas Garrett, Provision is made tor young and old, grave and W .if^ '>.^ gay. The subjects are well chosen, and the dia- logues are full oi life and sparkle. Excelsior Dialogues. By Phineas Garrett. Contains a wide variety of rflew and original dia- logues expressly prepared for this work by a corps of especially qualified writeis. JStlreka Entertainments. The weary^earch- er after material for entertainments will, upon ex- amination of this book, at once exclaim, " I have found it." Found just what is wanted lor use in Day-school, Sunday-school, at Church Socials, Teas, jind other Festivals, or for Patlor or Fire- side Amusement. Holiday Seleotions. For Readings and Recitations. By Sara S. Rice. The selections are specially adapted to Christmas, New Year's, St. Valentine's Day, Easter, Arbor Day, and Thanksgiving. ^ . Holiday Entertainments. By Charles C. Shoemaker. Contains many original exercises and other entertainments suitable, not only to the Christmas Holidays, but also to Easter, Thanks- giving, etc. Select Speeches for Declamation. 'By Jojin H. Becht«l. A volume especiaUy prepared for use by college men. A superior collection of short prose extracts from the leading orators and writers of all ages and nations. Itemperance Selactions. For Rsadings and Recitations. By John H. Bechtel. 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Much of the material was specially written, and all is eminently adapted to the pur- p»se m mind. School and Parlor Comedies. By li. L. C. Grififith. The plays differ w^idely in character, thus a fording an unusual variety. The scenery required is in no instance difficult, the situations are ah.ays ingenious, and the plots are such as command the attention begin'iing to the end. of an audience from the Monologues and Novelties. Bf B. L. C. Grifiith. This unique volume contains a dozen or more Monologues, a Shadow Pantomime, a Reci- tation with Lesson Help, a Play, and numerous other features, in ali of w>-ich are embodied many new and original ideas. How to Become a Public Speaker. By William Pittenger. This work shows in a simple and concise way how any person of ordinary per- severance and average intelligence may become a ready and effective public sjjeaker. THE CELEBRATED 100 CHOICE SELECTIONS 36 Different Numbers Paper Binding, Each, 30 Cents. One Hundred Choice Selections. 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