13. The triangle formed by joining the middle point of one of the non-parallel sides of a trapezium to the extremities of the opposite side is equal to half the trapezium.
PROP. XXXIX.—Theorem.
Equal triangles (BAC, BDC) on the same base (BC) and on the same side of it are between the same parallels.
Dem.—Join AD. Then if AD be not parallel to BC, let AE be parallel to it, and let it cut BD in E. Join EC. Now since the triangles BEC, BAC are on the same base BC, and between the same parallels BC, AE, they are equal [xxxvii.]; but the triangle BAC is equal to the triangle BDC (hyp.). Therefore (Axiom i.) the triangle BEC is equal to the triangle BDC—that is, a part equal to the whole which is absurd. Hence AD must be parallel to BC.
PROP. XL.—Theorem.
Equal triangles (ABC, DEF) on equal bases (BC, EF) which form parts of the same right line, and on the same side of the line, are between the same parallels.
Dem.—Join AD. If AD be not parallel to BF, let AG be parallel to it. Join GF. Now since the triangles GEF and ABC are on equal bases BC, EF, and between the same parallels BF, AG, they are equal [xxxviii.]; but the triangle DEF is equal to the triangle ABC (hyp.). Hence GEF is equal to DEF (Axiom i.)—that is, a part equal to the whole, which is absurd. Therefore AD must be parallel to BF.
Def.—The altitude of a triangle is the perpendicular from the vertex on the base.
Exercises.