48 



THE POPULAR EDUCATOR. 



But A c = B c ; and by Axiom 1 things which are equal to the 

 same thing are equal to one another. Therefore A c, A F, B c, 

 B F are all equal. Then, in the two triangles A c F, B c F, since 

 A c = c B, and c F is common, also base A F = base F B, there- 

 fore, by Euclid I. 4, the angle A c F = the angle B c F. 



Hence the angle A c B is bi- 

 sected by the line c F. (a) 



Again, in the two triangles A c G, 

 B c G, because side A c= side c B, and 

 side c G is common, also included 

 angle A c G has just been proved 

 equal to included angle B c G, there- 

 fore, by Euclid I. 8, 



Base A G = base G B. (0) 



Again, in the two triangles AGO, B G c, because side A G = side 

 G B, and side G c is common ; also base A c = base c B ; there- 

 fore, by Euclid 1. 4, included angle A G c = included angle B G c. 

 But these are adjacent angles which the straight line c G makes 

 by standing on the straight line AB. Therefore, by Def. 10, 

 each of them is a right angle. (7) 



This gives us at once the results of Props. IX., X., XI., and 

 XII. of Book I. 



PROPOSITION II. In the figure of Euclid I. 2 (Fig. 2), required 

 to draw from D a straight line D M N, cutting the circles in M and 

 N, such that M N, the part intercepted between them, may be 

 equa.l to A L or B c. 



Suppose D M N to bo thus drawn, then r> N, r> L, being radii 

 of the same circle, will be equal ; 

 but MN, by the supposition, is 

 equal to A L ; and, by Axiom 3, 

 if equals be taken from equals 

 the remainders are equal. There- 

 fore, if M N, A L be taken from 

 D N, D L, the remainders D M, 

 D A will be equal. Hence, if 

 from D, as centre, we describe 

 a circle, with D A as radius, cut- 

 ting the small circle in M and 

 joining D M, produce it to meet 

 the large circle in N, D M N will 

 be the line required ; for, since 

 D A, D M are radii of the same 

 circle, DA = D M (Def. 15) ; and, 

 since D L, D N are radii of the 

 same circle, D L = D N. Therefore, 



by Axiom 3, if D A, D M are taken from D L and D N, the re- 

 mainders A L, M N will be equal ; that is, the part of D N inter- 

 cepted between the circles will be equal to A L, and therefore 

 equal also to B c. Q. E. F. 



Corollary. It is obvious that there will be certain limits 

 beyond which the problem will be impossible. This will be the 

 case when the circle described with centre D and radius D A 

 does not cut the smaller circle. When it touches the smaller 

 circle, there will be only one possible position for r> M N, that of 

 passing through the point of contact. When it cuts there will 

 be two positions, as indicated in the figure by the dotted line. 

 PROPOSITION III. On a given base AB (Fig. 3), to describe an 



isosceles triangle 

 ABC, such that 

 each of the sides 

 A c, B c may be 

 four times the 

 base A B. 



Produce A B 

 both ways indefi- 

 nitely to E and F, 

 and from B E, the 

 greater in the side 

 remote from A, cut 

 off B G equal to 



three times B A (I. 3), and from A F on the side remote from 

 B cut off A H equal to three times A B (I. 3), then A a will be 

 equal to four times A B, and B H will be equal to four times B A. 

 From centre B, at distance B H, describe a circle (Post. 3), and 

 from centre A, at distance A G, describe another circle, and let 

 these two circles cut in c. Join A c, B c, then ABC shall be the 



Fig. 2. 



Fig. 3. 



triangle required. For since A c, A G are radii of the same circle, 

 A c = A G ; but A G = 4 A B, therefore A c = 4 A B. Similarly 

 = BH=4AB; therefore A c = B c, and the triangle ABO 

 is isosceles, and each of its sides equal to four times the base. 

 Q. E. F. 



PROPOSITION TV. If ABC (Fig. 4) be an isosceles triangle, 

 vertex A, and two points H, K be taken in the sides A B, A c, such 

 that A H = A K, and if B K, c H meet in F, then A F bisects the 

 angle BAG. 



For in the triangles c A H, B A K, since A c = A B, and 

 A H = A K, also the included angle c A B is common to the two 

 triangles, therefore the base B K is equal to the base c H (I. 8). 

 Again, in the triangles A c H, A B K, because A c is equal to A B and 

 C H to B K, and also the 

 base A H to the base A E, 

 therefore the included angle 

 A c H is equal to the included 

 angle A B K (I. 4). But the 

 whole angle A c B = whole 

 A B c (I. 5), and, by Axiom 3, 

 if equals be taken from 

 equals the remainders are 

 equal ; therefore if A c H, 

 A B K be taken from A c B, 

 ABC, the remainders F c B, 



F B c will be equal, and consequently the side F c = side 

 FB (I. 6). Then, in the two triangles CAF, BAF, because 

 A c A B and A F is common, also base F c base F B, there- 

 fore also included angle F A c is equal to included angle FAB, 

 that is, A F bisects the angle BAG. Q. E. D. 



PROPOSITION V. If A, B (Fig. 5) be two points on opposite sides 

 of a line c D, required to find in c D a point E, such that the 

 angle A E c may equal the angle EEC. 



By means of Prop. I. wo may draw A F perpendicular to c D. 

 Produce AF to G (Post. 2), and make F G FA (I. 3). Join 

 G B, and produce it to cut c D in E, then E is the point required. 

 Join A E. Then, since A F F G, and F E is common to the 

 two triangles A F E, G F E, also the 

 included right angle A F E is equal 

 to the included right angle G F E, 

 therefore the base A E is equal to 

 the base G E (I. 4). Again, because 

 A E = E G, and F E is common, also 

 base A F = base F G, therefore in- C 

 eluded angle A E F - included angle 

 G E F (I. 8), therefore c D bisects the 

 angle A EB. Q. E. F. 



In the present paper we have 

 used Book I. 1 8. In our next 

 paper we shall use Book I. 1 16, 

 and give solutions of the following 

 propositions : 



PROPOSITION VI. In the figure of Euc. I. 5, if G o at right 

 angles to A G cut A H produced in o, H being the intersection 

 of B G, F c, then o F shall be perpendicular to A F. 



PROPOSITION VII. If A c, the side of an isosceles triangle 

 A B c, be bisected in D, and B D produced to E, so that D E = 

 r> B, then if A E be joined, the angle A E D shall be equal to the 

 angle DEC. 



PROPOSITION VIII. If A B be two points on the same side 

 of a given line c D, find in c D a point E, such that the angle 

 AEG may be equal to the angle BED. 



PROPOSITION IX. Given two straight lines A B, A c, intersecting 

 in A, and another straight line D of limited length : required to 

 form a right-angled triangle of which the base shall coincide 

 with A c, one side shall coincide with A B, and the other side be 

 equal to D. 



PROPOSITION X. In an isosceles triangle A B c, if A L be 

 drawn from vertex A perpendicular to the base B c, and if A L 

 be produced to M so that L M = L A, then shall B L be equal to 



BA. 



PROPOSITION XI. If in any triangle the sides A B, A c be 

 bisected in L,M, and L o, M o be drawn at right angles to A B, A c, 

 meeting in o, then o N, drawn perpendicular to B c, will bisect 



BC. 



PROPOSITION XII. In the figure of Eno. I. 9, if with centre 

 A and radius A F, a circle be described cutting A B, A C in I. M. 

 then shall E L be equal to B H. 



Fig. 5. 



