Observation.—By the second method of proof the subdivision of the demonstration into cases is avoided. It is easy to see that either of the two parallelograms ABCD, EBCF can be divided into parts and rearranged so as to make it congruent with the other. This Proposition affords the first instance in the Elements in which equality which is not congruence occurs. This equality is expressed algebraically by the symbol =, while congruence is denoted by ≡, called also the symbol of identity. Figures that are congruent are said to be identically equal.
PROP. XXXVI.—Theorem.
Parallelograms (BD, FH) on equal bases (BC, FG) and between the same parallels are equal.
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.