IMAGE EVALUATION TEST TARGET (MT-3) / O w 1.0 If'** I I.I 1.25 2.5 S ^ 1^ 1^ 12.0 1.8 1-4 IIIIII.6 V] <^ /a <3 '>> 7 Wl W '/ i\ <'> CIHM/ICMH Microfiche Series. CIHM/ICMH Collection de microfiches. Canadian Institute for Historical Microreproductions Institut Canadian de microreproductions historiques 1980 Technical Notes / Notes techniques The Institute has attempted to obtain the best original copy available for filming. Physical features of this copy which may alter any of the images in the reproduction are checked below. m D D D Coloured covers/ Couvertures de couleur Coloured maps/ Cartes gdographiques en couleur Pages discoloured, stained or foxed/ Pages d6colordes. tachetdes ou piqu6es Tight binding (may cause shadows or distortion along interior margin)/ Reliure serrd (peut causer de I'ombre ou de la distortion le long de la marge int^rieure) L'Institut a microfilm^ le meilleur exemplaire qu'il lui a 6t6 possible de se procurer. Certains ddfauts susceptibles de nuire d la quality de la reproduction sont notds ci-dessous. D D m D Coloured pages/ Pages de couleur Coloured plates/ Planches en couleur Show through/ Transparence Pages damaged/ Pages endommagdes The pos oft film The con or t app The filrr insl Ma in ( upf bot foil D Additional comments/ Commentaires suppldmentaires Bibliographic Notes / Notes bibliographiques n Only edition available/ Seule Edition disponible D Pagination incorrect/ Erreurs de pagination □ Bound with other material/ Relid avec d'autres documents Cover title missing/ Le titre de couverture manque n Pages missing/ Des pages manquent Maps missing/ Des cartes gdographiques manquent n Plates missing/ Des planches manquent D Additional comments/ Commentaires suppldmentaires The images appearing here are the best quality possible considering the condition and legibility of the original copy and in keeping with the filming contract specifications. Les images suivantes ont 6t6 reproduites avec le plus grand soin, compte tenu de la condition et de la nettet6 de I'exemplaire film6, et en conformity avec les conditions du contrat de filmage. The last recorded frame on each microfiche shall contain the symbol —»► (meaning CONTINUED"), or the symbol V (meaning "END"), whichever applies. Un des symboles suivants apparaTtra sur la der- nidre Image de cheque microfiche, selon le cas: le symbols — ► signifie "A SUIVRE", le symbole y signifie "FIN". The original copy was borrowed from, and filmed with, the kind consent of the following institution: National Library of Canada L'exemplaire film6 fut reproduit grfice d la g6n6rosit6 de I'dtabiissement prdteur suivant : Bibliothdque nationale du Canada Maps or plates too large to be entirely included in one exposure are filmed beginning in the upper left hand corner, left to right and top to bottom, as many frames as required. The following diagrams illustrate the method: Les cartes ou les planches trop grandes pour dtre reproduites en un seul clich6 sont filmdes d partir de I'angle sup6rieure gauche, de gauche d droite et de haut en bas, en prenant le nombre d'images n6cessaire. Le diagramme suivant illustre la mdthode : 1 2 3 1 2 3 4 5 6 EXTRAORDINARY TRICAL DISCOVERY TRISEGTION OF ANY RECTILINEAL ANGLE — BY ELEMENTARY GEOMETRY « AND SOLUTIONS OF OTHER PROBLEMS Considered impomble except by aid of the higher Geometry, — BY — AjYDREW' DOYLE. '--'-'^'^S^ ^?2/»t-V-t OTTAWA: A. BUREAU, PRINTEB. SPARKS STREET 1880 n^ aJ ,.^>-\_- Eiitoi'cd ftccovding to Act of Parliament of Canada, in the year 1880, ^y AndrbW Doylb, in the office of the Miniator of Agriculture. \ Trisection of a Rectilineal Angle. Let B O C be the aiiujlo to be triwected. At the point O, make (23.1) the aii.^les BOD and C O J*^, each equal to the angle B O C. Take O as centre with any radius OB, and describe the arc DBCE; join D B, B 0, C E, and D K. Produce I) B, and E C to meet in T; join O T. By 4.1., the angle O D B is equal to the angle O K C, and (5.1) the angle O D E is equal to the angle O E D ; take away the two latter angles and the remaining angles E D T and I) E T are equal : therefore (0.1) I) T is equal to E T. In the triangles O D T and O E T, the three sides of the one are respectively equal to the three sides of the other, by (8.1) they are equal in every respect; therefore the angle D T O is equal to the angle E T O, then the line O T i)isecting the vertical angle of an isosceles triangle, bisects D E and B C perpendicularly. It also bisects the arc BO; it is evident by 4.1 that D B, B C, and C E are equal to one another. Draw a d parallel to B C. so that a d shall be equal a B and d C. This is done by bisecting the angles T B C and T C B by lines meeting the sides B T and C T in a and d, when the triangle is isosceles, or equilateral and proved hy (29.1, 6.1, and 4.1). Find the diamater of a circle circumscribing the trapizium B a (Z C, and make t L equal to it ; join a L and d L, cutting the arc B C in the points r and I : these are the trisecting points of the arc B C which is t measure of the given angle B O C. Draw the lines Orb ard O I c, and the three angles 6 c, 6 O B, and C O c are equal and each equal to one third of the given angle B C. . H E FIRST DEMONSTEATION. Through the point B, di'aw B G paralled to O / or O c, and meeting 1) E in G ; draw also B P parallel to O r or O 6; make (3.1) B A equal to B G ; through G draw G H parallel to B O, and through the point A draw A H parallel to B G ; join B H, and B H passew through the point P. We have now, the straight line B P and the line B H, having the two points B and P in common ; therefore (Prop. 2, Legendre,) they coin cide throughout the whole, or they are in one and the same straight line. A B is parallel to G H, and A H is parallel to B G ; therefore A B G H is a parallelogram, having its op- {DObite sides equal and parallel, and A B is equal to B G )y construction ; therefore A B G H is a rhombus and the diagonal B H bisects its opposite angles ; therefore the angle A B P is equal to the angle GBP. B G is parallel to O / and B O a line meeting them, by (29.1), the angle A B G is equal to the angle BO/; for the same reason, the angle A B P is equal to the angle B O r ; therefore the remaining angle G B P is equal to the angle r O I but GBP and A B P are equal ; B or and roe similarly it can be proved that the angle I O C is equal to the angle r O I] but things which are equal to the same thing, are equal to one another; therefore the angles B O r, r O /, and I O C are equal, and the angle B O C is trisected. SECOND SEMONSTBATION. Suppose L C and L B to be joined, it is easily proved that L C is equal to L B, then by 5.1, the angle L B C is equal to the angle L C B, and the angle B C T is equal to the angle C B T ; therefore the whole angle 6 Jj C d in equal to the whole angle L B a. In the triangles d C Jj and a Blj, a B and B L are equal to d and C L, and their contained angles equal, by (4.1), a L Ih equal to d L, and the angle B a L equal to the angle C d L, by (13.1), the angle r a b 'i» equal to the angle / d c. In the triangle r L O and I L O, the three sides of the one are respectively equal to the three sides of the other, by (8.1), O r L is equal to the angle O / L, and the angle r O L eqiuil to the angle / O L; by (15.1) the angle orb and die are equa but it has been proved that the angle barm equal lo c d I; therefore (32.1), the two triangles are equiangular, and a r equal \o d i\ therefore, by (2().l), a 6 is equal to c d^ but a B is equal d C ; therefore b B is equal to c C. Then, in the triangles 6 B O and c C O, we have b B and B O equal to c C and C O and their contained angles equal, by (4.1), the angle B O 6 is equal to the angle C O c. Then, as in the first demonstration it may be proved that each of these angles^ is equal to the angle 6 0^; therefore the three angles B O /, r O /, and I O C, are equal, and the angle B O C is trisected. THIED DEMONSTRATION. Join B I and C r, cutting O / and O r in n and m. If on O B we describe a semicircle, it passes through the point m; therefore (31.3), the angles at m are right angles, and (3.3) B I is bisected perpendicularly by O m ; hence B m and m O are equal to O m and m I and the angles at m equal, by (4.1), the angle B O m is equal to the angle I O m. This may also be proved by letting fall a perpendicular from O, on B Z and this per- pendicular always falls on the point m ; then we have two straight lines which coinside in part they must coinside throughout the whole (Legendre, Prop. 2). It may be similarly proved that r is bisected perpendi- cularly by O c in w, and that the angle C O n is equal to the angle r O w ; therefore the three angles, B O r, r O /, and / O C, are equal ; therefore the angle B O C is trisected. POUETH DEMONSTEATION. In O T, take any point, as L, and with L O as radius describe the circle O U V S. Through Y draw V X parallel to S O, meeting the circle in X ; and through u draw u X parallel to V O. Because S O is parallel to V X and V O meeting them, (29.1; the angle S O V is equal to the angle O V X ; for the same reason, the angle O V X is ^^qual to the angle V X U ; but V X U is equal to V O U ; therefore the angle S O V is equal to the angle V O U. Similarly the angle C O U may be proved equal to V O U ; there- fore S O V=V O U=U O C, and the angle B O C is trisected. The lines V X and U X, in the trisection of ever}^ angle, always meet the circumference of the circle in the same point. This evidently, can never happen but when the three arcs B r, v I, and I C are equal. ANDREW DOYLE. w 8 The following problems also have been considered by all mathematicians to be beyond the range of elemen- tary geometry. 1. Let BAG (Fig. 2) be any angle ; take any point D in A B and let fall the perpendicular D E ; through D, draw D P parallel to AC; it is required to draw the line A N O P, so that N P may be equal to twicQ AD. Trisect the angle B A C by A P and N P is the line required. Bisect N P in O, and join O D. red to ^iv, 80 that the angle A CI H may be equal to twice the angle AHG. Trisect B A C by A P ; take any point G and draw G II parallel to A P, meeting A C in II. AN DEE W DOYLE. d 2. Let C A G be any angle, (Fig. 2) it is requii raw a line G H cutting the legs of the angle, sc ll In the next Edition will appear the method of finding two geometrican means between two straight lines, and also the dupli- cation of the cube. ^ ■^ MMMMi