Let the two triangles A B C, A D C, have the side A B of the one equal to the side A D of the other; the side B C of the one equal to the side D C of the other, and the third side likewise equal; then will the two triangles be equal in all their parts. For place the two triangles together by their longest side, and join the opposite vertices B and D by a straight line. Because the side A B is equal to the side A D, the triangle B A D is isosceles, and the angles A B D, A D B, opposite the equal sides are equal. Because the side B C is equal to the side D C, the triangle B C D is isosceles, and the angles C B D, C D B, opposite the equal sides are equal. If to the angle A B D we add the angle D B C, we shall have the angle A B C. And if to the equal of A B D, that is, A B D, we add the equal of D B C, that is, B D C, we shall have the angle A D C. Therefore the angle A B C is equal to the angle A D C. Then the two triangles A B C, A D C, have two sides, and the included angle of the one equal to two sides and the included angle of the other, each to each, and are equal in all their parts; that is, the three angles of the one are equal to the three angles of the other, and their areas are equal. [Illustration] PROPOSITION XX. THEOREM. DEMONSTRATION. We wish to prove that _An angle at the circumference is measured by half the arc on which it stands._ Let B A D be an angle whose vertex is in the circumference of the circle whose centre is C; then will it be measured by half the arc B D. For through the centre draw the diameter A E, and join the points C and B. The exterior angle E C B is equal to the sum of the angles B and B A C. Because the sides C A, C B, are radii of the circle, they are equal, the triangle is isosceles, the angles B and B A C opposite the equal sides are equal, and the angle B A C is half of both. Then, because the angle B A C is half of B and B A C, it must be half of their equal E C B. But E C B, being at the centre, is measured by B E; then half of it, or B A C, must be measured by half B E. In like manner, it may be proved that the angle C A D is measured by half E D. Then, because B A C is measured by half B E, and C A D by half E D, the whole angle B A D must be measured by half the whole arc B D. SECOND CASE. Suppose the angle were wholly on one side of the centre, as F A B. Draw the diameter A E and the radius B C as before. Prove that the angle B A E is measured by half the arc B E. Draw another radius from C to F, and prove that F A E is measured by half the arc F E. Then, because the angle F A E is measured by half the arc F E, and the angle B A E is measured by half the arc B E,