Dem.—Let the same construction be made, then we have AC : CD :: triangle ACB : BCD [i.], and EC : CB :: triangle DCE : BCD [i.]; but AC : CD :: EC : CB (hyp.). Therefore the triangle Hence the triangle ACB = DCE [V. ix.]—that is, the triangles are equal. 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. PROP. XVII.—Theorem 1. If three right lines (A, B, C) be proportional, the rectangle (A.C) contained by the extremes is equal to the square (B2) of the mean. 2. If the rectangle contained by the extremes of three right lines be equal to the square of the mean, the three lines are proportional. Dem.—1. Assume a line D = B; then because A : B :: B : C, we have A : B :: D : C. Therefore [xvi.] AC = BD; but BD = B2. Therefore AC = B2; that is, the rectangle contained by the extremes is equal to the square of the mean. 2. The same construction being made, since AC = B2, we have A.C = B.D; therefore A : B :: D : C; but D = B. Hence A : B :: B : C; that is, the three lines are proportionals. This Proposition may be inferred as a Cor. to the last, which is one of the fundamental Propositions in Mathematics. Exercises.