Dem.—Join BE, CH. Now since FH is a parallelogram, FG is equal to EH [xxxiv.]; but BC is equal to FG (hyp.); therefore BC is equal to EH (Axiom i.). Hence BE, CH, which join their adjacent extremities, are equal and parallel; therefore BH is a parallelogram. Again, since the parallelograms BD, BH are on the same base BC, and between the same parallels BC, AH, they are equal [xxxv.]. In like manner, since the parallelograms HB, HF are on the same base EH, and between the same parallels EH, BG, they are equal. Hence BD and FH are each equal to BH. Therefore (Axiom i.) BD is equal to FH. Exercise.—Prove this Proposition without joining BE, CH. PROP. XXXVII.—Theorem. Triangles (ABC, DBC) on the same base (BC) and between the same parallels (AD, BC) are equal. Dem.—Produce AD both ways. Draw BE parallel to AC, and CF parallel to BD [xxxi.] Then the figures AEBC, DBCF are parallelograms; and since they are on the same base BC, and between the same parallels BC, EF they are equal [xxxv.]. Again, the triangle ABC is half the parallelogram AEBC [xxxiv.], because the diagonal AB bisects it. In like manner the triangle DBC is half the parallelogram DBCF, because the diagonal DC bisects it, and halves of equal things are equal (Axiom vii.). Therefore the triangle ABC is equal to the triangle DBC. Exercises. 1. If two equal triangles be on the same base, but on opposite sides, the right line joining their vertices is bisected by the base. 2. Construct a triangle equal in area to a given quadrilateral figure. 3. Construct a triangle equal in area to a given rectilineal figure. 4. Construct a lozenge equal to a given parallelogram, and having a given side of the parallelogram for base. 5. Given the base and the area of a triangle, find the locus of the vertex. 6. If through a point O, in the production of the diagonal AC of a parallelogram ABCD, any right line be drawn cutting the sides AB, BC in the points E, F, and ED, FD be joined, the triangle EFD is less than half the parallelogram. PROP. XXXVIII.—Theorem. Two triangles on equal bases and between the same parallels are equal. Dem.—By a construction similar to the last, we see that the triangles are the halves of parallelograms, on equal bases, and between the same parallels. Hence they are the halves of equal parallelograms [xxxvi.]. Therefore they are equal to one another. Exercises. 1. Every median of a triangle bisects the triangle. 2. If two triangles have two sides of one respectively equal to two sides of the other, and the contained angles supplemental, their areas are equal. 3. If the base of a triangle be divided into any number of equal parts, right lines drawn from the vertex to the points of division will divide the whole triangle into as many equal parts.