IMO 1985 SL 20

Origin: GBR

Problem

A circle whose center is on the side ED of the cyclic quadrilateral BCDE touches the other three sides. Prove that EB+CD = ED.

Solution

Let O be the center of the circle touching the three sides of BCDE and let F \in(ED) be the point such that EF = EB. Then \angleEFB = 90◦− \angleE/2 = \angleC/2 = \angleOCB, which implies that B, C, F, O lie on a circle. It follows that \angleDFC = \angleOBC = \angleB/2 = 90◦−\angleD/2 and consequently \angleDCF = \angleDFC. Hence ED = EF + FD = EB + CD. Second solution. Let r be the radius of the small circle and let M, N be the points of tangency of the circle with BE and CD respectively. Then EM = r cot E, DN = r cot D, MB = r cot(\angleB/2) = r tan(\angleD/2), NC = r tan(\angleE/2), and ED = EO + OD = r/sin D + r/sin E. The statement follows from the identity cot x + tan(x/2) = 1/sin x.