270 



THE POPULAR EDUCATOR 



EXERCISES IN EUCLID. IV. 



PROPOSITION XXI. Given two straight lines A B, A c (Fig. 

 20) meeting in A, and another line D E, of limited length. Required 

 to describe an isosceles triangle A L M, such that A L may coincide 

 with A B, and A M with A c, and L M may be equal to D E. 



Take two points p, Q in A B, A c, such that A P = A Q. Join 

 P Q, and produce p Q to R, so that 

 p R may be equal to D E. At the 

 point R in the straight line P R 

 make the angle p R M equal to the 

 angle APQ (Euc. I. 23), and let 

 R M cut A c in M ; then if A L be 

 taken in A B equal to A M, A L M 

 will be the triangle required. 



For because the angles at the 

 base of an isosceles triangle are equal (Euc. I. 5), and that A Q p 

 = R Q M (Euc. I. 15), and that A p Q = Q R ji by construction, 

 therefore angle M R Q = M Q R; therefore, by Euc. L 6,'M = 

 M Q. But since A M = A L, and A Q = A p, therefore, by Axiom 

 III., M Q = P L ; therefore also R M = p L (Euc. 1. 28). Also, since 

 angle A P Q = Q R M, therefore A L is parallel to R M (Euc. I. 

 27); therefore it M is equal and parallel to PL. Join PM, then 

 the angle R M p = angle M p L (Euc. I. 29) ; also side R M = 

 side P L, and side M P is common, therefore base R P = base M L 

 i.e., M L = D E (Euc. I. 8), and A M = A L ; therefore A M L is 

 the triangle required. Q. E. F. 



PROPOSITION XXII. If ABC (Fig. 21) be an isosoceles triangle 

 of vertex A, with B c the base pro- 

 duced to D, and if from centre c, at 

 distance c A, a circle be described 

 cutting B A produced in E, then 

 exterior angle BCD shall be equal 

 to three times angle ABC or A c B. 

 For since c A, c E are radii of the 

 same circle, angle c A E = angle 

 c E A (Euc. 1. 5) ; but exterior angle 



Fig. 21. 



Fipr. '2. 



c A E equals two interior and opposite angles ABC and A c B = 

 twice angle A c B (Euc. I. 32) ; therefore angle c E A equals twice 

 angle A B c or A c B. But exterior angle BCD equals two interior 

 and opposite angles ABC and A E c (Euc. I. 32) equals three 

 times angle A B c or A c B. Q. E. D. 



PROPOSITION XXIII. In the figure of Euc. I. 5, draw c L 

 (Fig. 22) at right angles to c B, meeting B A 

 produced in L, and prove A L A c. Since 

 the three angles of a triangle are together 

 equal to two right angles (Euc. I. 32), and 

 the angle ACL is a right angle, therefore 

 the angles c B I,, B L c of the triangle B c L 

 are together equal to a right angle. But the 

 angle A c B is equal to the angle A B c, there- 

 fore angle A c B with A L c is equal to a 

 right angle. But A c B with A c L is equal 

 to B c L, that is, to a right angle ; there- 

 fore A C B with A C L is equal to A C B with 

 AL C ; therefore, taking away A c B, we have 

 remainder A c L = remainder A L c r therefore 

 AC = AL (Euc. I. 6). Q. E. D. 

 PROPOSITION XXIV. If in the figure of Euclid I. 5 the 

 angles F B c, B c G (Fig. 23) be bisected by lines c o, B o, meet- 

 ing in o, then o A shall bisect the angle B A c. Draw o L, o M, 

 o N perpendicular to G c, c B, B r, then in the two triangles 

 o L c, o N c, because the right angle o L c = right angle o N c, 

 and angle L c o = angle N c O, each being by construction half 

 the angle L c N, also side ^ 



O C is common, therefore 

 triangle o L c = triangle 

 ONC (Euc. I. 26); there- 

 fore c L = c N, and o L 

 o N. Similarly, B M B N, 

 and o M = o N ; but since 

 O N B, o N C are right angles, 

 and angles o c N, o B N are 

 equal, being halves of the 

 equal angles L c N, MEN, and side o N is common (Euc. I. 5), 

 therefore the triangles o N B, o N c are equal; therefore c N = N B. 

 But L c c N, and M B = B N; therefore L c = B M. But AC = 

 A B ; therefore A L A M, and o L, o M are equal, each being equal 



to o N. Therefore in the two triangles o A L, o A M, since O L, 

 L A are equal to o M, M A, each to each, and o A is common, 

 therefore the triangles are equal (Euc. I. 8) ; therefore angle 

 O A L = angle O A M. Q. E. D. 



Note. It will be observed in the figure that o N coincide?, 

 with o A, but this is not assumed in the proof. This proposition 

 will be proved in a future article for any triangle not isosceles. 

 PROPOSITION XXV. In the figure of Euo. I. I, if the circles 

 cut again in F (Fig. 24) and c A produced cut the circle in H, 

 then H F is equal to A B. Join 

 F A, F B, F c ; then A B, A c, A F, 

 AH, B c, B F are equal, being 

 radii of equal circles. Hence 

 ABC, A B F are equilateral tri- 

 angles, and therefore also equi- 

 angular (Euc. I. 5, Cor.) ; but, 

 by Euc. I. 32, the angles of a 

 triangle are together equal to 

 two right angles ; therefore 

 each angle of an equiangular 

 triangle is equal to one-third 

 of two right angles i.e., each of the angles c A B, B A F is equal 

 to one-third of two right angles. Therefore angle OAF equals 

 two-thirds of two right angles ; hence, by Euclid I. 13, angle 

 F A H is equal to one-third of two right angles ; therefore .angles 

 A H F, A F H are together equal to two-thirds of two right angles 

 (Euo. I. 3), but they are equal, since AH = AF; therefore 

 each of them is equal to one-third of a right angle, hence the 

 triangle A H F is equiangular, and therefore also equilateral 

 (Euc. I. 6, Cor.) ; therefore H F is equal to H A, or equal to 

 AB. Q. E. D. 



PROPOSITION XXVI. If in a triangle ABC (Fig. 25), B c be 

 bisected in G, and A G joined, and tho angle B A G b;; equal to the 

 angle GAG, then B A shall be equal to c A. For if A B be not equal 

 to A c, one of them must be greater. Let A B be tho greater ; from 

 A B cut off A H equal to A c. Join c H, o H ; then because A H = 

 A c, and AG is common, also included angle GA W = included angle 

 G A c, therefore base G H = G c (Euc. 

 I. 4). But G c = G B, therefore G H = 

 G B, and angle G H B = angle G B H (Euc. 

 1.5). Also since G H G c, angle G c H 

 = angle GHC (Euc. I. 5). Therefore 

 angles GHC and G H B are together 

 equal to G B H and G c H together 

 i.e., angle B H c equals sum of B c H and 

 c B H ; but the three angles of the tri- 

 angle are together equal to two right 



Fig. 25. 



I angles, therefore angle B H C, which is equal to half the sum of the 

 | three angles, is a right angle. Therefore, also, c H Ais a right angle 

 ! by Euc. I. 13 ; and since A H = A c by supposition, angle A c H 

 I = angle A H c (Euc. I. 5) ; therefore also ACH is a right angle 

 ! i.e., two angles A c H, A H c of the triangle A c H are equal to 

 , two right angles, which is impossible by Euc. I. 17 ; hence AH 



is not equal to A c ; and similarly for any other point, except B. 

 , Hence A B is equal to A c. 



Note. This kind of proof is called " Eeductio ad absurdum." 

 PROPOSITION XXVII. Given two straight lines, A B, A c (Fig. 

 : 26), meeting in A, and another straight line, D, of limited length. 

 , Required to form a right-angled triangle, of which the base shall 

 i coincide with A C ; one side shall | 

 i coincide with A B, and the other side E 

 ! be equal to D. 



From A draw A E at right a.ngles 



to AB (Euc. I. 11), and from A E cut 



off A F, equal to (Euc. I. 3) ; from 



F draw F G at right angles to A E, 



cutting AC in G (Euc. I. 11) ; from 



G draw G H perpendicular toAB (Euo. 



Fig. 23. 



1. 12) ; then will A H G be the triangle 

 required. For since G H A, H A F, A F G are right angles, A H G P 

 is a rectangular parallelogram ; hence in the two triangles F A G, 

 A G H, because the angle F A G is equal to alternate angle A G H 

 (Euc. I. 29), and right angle AFG equal to right angle AHG, 

 also side A G is common, therefore the triangles are equal in 

 every respect (Euc. I. 26). Therefore side A F = G H, but A P 

 = D ; therefore G H = D, and A H coincides with A B, and A G 

 with A c, and A H G is a right anglo ; hence A n o is the trianglo 

 required. <$. B. F. 



