This Proposition might have been appended as a Cor. to the preceding, since the triangles are the halves of equiangular parallelograms, or it may be proved by joining AE, and showing that it is parallel to BD.
PROP. XVI.—Theorem.
1. If four right lines (AB, CD, E, F) be proportional, the rectangle (AB.F) contained by the extremes is equal to the rectangle (CD.E) contained by the means.
2. If the rectangle contained by the extremes of four right lines be equal to the rectangle contained by the means, the four lines are proportional.
Dem.—1. Erect AH, CI at right angles to AB and CD, and equal to F and E respectively, and complete the rectangles. Then because AB : CD :: E : F (hyp.), and that E is equal to CI, and F to AH (const.), we have AB : CD :: CI : AH. Hence the parallelograms AG, CK are equiangular, and have the sides about their equal angles reciprocally proportional. Therefore they are [xiv.] equal; but since AH is equal to F, AG is equal to the rectangle AB.F. In like manner, CK is equal to the rectangle CD.E. Hence AB.F = CD.E; that is, the rectangle contained by the extremes is equal to the rectangle contained by the means.
2. If AB.F = CD.E, to prove AB : CD :: E : F.
The same construction being made, because AB.F = CD.E, and that F is equal to AH, and E to CI, we have the parallelogram AG = CK; and since these parallelograms are equiangular, the sides about their equal angles are reciprocally proportional. Therefore
Or thus: Place the four lines in a concurrent position so that the extremes may form one continuous line, and the means another. Let the four lines so placed be AO, BO, OD, OC. Join AB, CD. Then because AO : OB :: OD : OC, and the angle AOB = DOC, the triangles AOB, COD are equiangular. Hence the four points A, B, C, D are concyclic. Therefore [III. xxxv.] AO.OC = BO.OD.