and because the angle BAC is equal to the angle BCA, therefore the side BA is equal to the side BC: but it has been proved that the side AB is equal to the side AC; therefore the side AC is equal to the side BC (ax. 1). Wherefore the equiangular triangle ABC is equilateral. Q.E.D. 8. Is the equality of the triangles in all respects proved ir this proposition? If not, how would you prove it? The equality of the triangles in all respects is not proved: it can be proved by 1. 4. If two circles cut each other, the line joining their points of intersection is perpendicular to the line joining their centres. Let the two circles CDE, CDF intersect at the points C, D, and let CD be joined, also A, B, the centres of the circles. Then CD shall be perpendicular to AB. Let G be the point of intersection of AB, CD, and draw AC, AD, BC, BD (post. 1). Proof-In the triangles CAB, DAB, the side CA is equal to the side DA (def. 15), the side AB is common to the two triangles, and the base BC is equal to the base BD (def. 15), therefore the angle CAB is equal to the angle DAB (1. 8). Again, in the triangles CAG, DAG, the two sides CÀ, AG art equal to the two sides DA, AG, each to each, and the angle CAG has been proved equal to the angle DAG, therefore the base CG is equal to the base DG. (1. 4), therefore AG is perpendicular to CD (def. 10). From every point of a given line, the lines drawn to each of two given points on opposite sides of the lines are equal: prove that the line joining the given points will be bisected by the given line at right angles. Let AB be the given line, and C, D the given points. Draw the lines FC, FD, GC, and GD from any two points F, G in AB, and join CD. Then, if FC be equal to FD, and GC equal to GD, Proof. In the triangles CGF, DGF, and the base FC is equal to the base FD (hyp.), therefore the angle CGF is equal to the angle DGF (1. 8). Again, in the triangles CGE, DGE, the two sides CG, GE are equal to the two sides DG, GE, each to each, and the angle CGE has been proved equal to the angle DGE, therefore the base CE is equal to the base DE (1. 4), and the angle CEG is equal to the angle DEG; and these are adjacent angles; therefore AB is at right angles to CD (def. 10). Wherefore CD is bisected at right angles by AB. Q.E.D. What is the hypothesis of this proposition? The hypothesis is that the three sides of the one triangle are equal to the three sides of the other triangle, each to each. 12. Explain why it is necessary to stipulate that the given line must be of unlimited length. If the line were of limited length, the given point might be situated so far from it. that the circle would fall without it altogether, or only cut it in one point. If in a triangle the perpendicular from the vertex on the base bisect the base, the triangle is isosceles. In the triangle ABC let the perpendicular AD bisect the base BC. Then the triangle ABC shall be isosceles. B D Proof. In the triangles ADB, ADC, the side DA is common to the two triangles, and the included angle BDA is equal to the included angle CDA (ax. 11), therefore the base AB is equal to the base AC (1. 4). Wherefore the triangle ABC is isosceles. Q.E.D. 13. If the angles are CBA, ABD, and BE bisects ABC, and BF bisects ABD, then FBE is a right angle. Proof. The angles ABC, ABD are together equal to two right angles (1. 13), the angle ABE is half the angle ABC, and the angle ABF is half the angle ABD (hyp.); therefore the angles ABE and ABF, that is, the angle EBF, are together equal to half the angles ABC and ABD; wherefore the angle EBF is equal to the half of two right angles; therefore the angle EBF is a right angle. Q.E.D. Show that the vertex of an isosceles triangle, and the intersections of the bisections of the interior and exterior angles at the base, are in the same straight line. Let ABC be an isosceles triangle, and let BD and CD bisect the angles at the base; also let EB bisect the exterior angle ABF. Join AD (post. 1), and produce it to meet EB produced in G (post. 2); join GC, and produce it to H. Then HC shall bisect the exterior angle ACK. Proof. The angle ABC is equal to the angle ACB (1.5); the angle DBC is half the angle ABC (hyp.), and the angle DCB is half the angle ACB; therefore the angle DBC is equal to the angle DCB; wherefore the side DB is equal to the side DC (1. 6). In the triangles BAD, CAD, the two sides AD, DB are equal to the two sides AD, DC, each to each, and the base AB is equal to the base AC (hyp.); therefore the angle ADB is equal to the angle ADC (1. 8). The angles ADB, BDG are together equal to two right angles (I. 13), and the angles ADC, CDG are together equal to two right angles; therefore the angle BDG is equal to the angle CDG (ax. 3). In the triangles BDG, CDG, the two sides BD, DG are equal to the two sides CD, DG, each to each, and the angle BDG is equal to the angle CDG; therefore the base BG is equal to the base CG (1. 4). The angle ACB is equal to the angle ABC (1. 5), and the angle GCB is equal to the angle GBC (1.5); therefore the whole angle ACG is equal to the whole angle ABG (ax. 2). The angles ACG, ACH are together equal to two right angles (1. 13), and the angles ABG, ABE are together equal to two right angles; therefore the angle ACH is equal to the angle ABE (ax. 3). The angle ABE is half the exterior angle ABF, and the exterior angle ACK is equal to the exterior angle ABF, the angle ACB being equal to the angle ABC; therefore the angle ACH is half the exterior angle ACK. Wherefore HC bisects the exterior angle ACK. Q.E.D. 15. If the lines are AEB, CED, and FE, GE bisect the angles AEC, DEB; then FEG is a straight line. F E B Proof. The angle FEC is half the angle AEC (hyp.), |