HC = twice the △ HBE, and the
CD = twice the △ BDE. Therefore the △ HBE = BDE, and [I. xxxix.] HD is parallel to BE. Hence
2. May be proved by reversing this demonstration.
Another demonstration of this Proposition may be got by producing the lines HA and DG to meet in I. Then [I. xliii.] the points I, C, F are collinear, and the Proposition is evident.
PROP. XV.—Theorem.
1. Two triangles equal in area (ACB, DCE), which have one angle (C) in one equal to one angle (C) in the other, have the sides about these angles reciprocally proportional.
2. Two triangles, which have one angle in one equal to one angle in the other, and the sides about these angles reciprocally proportional, are equal in area.