1985 IMO Problems/Problem 1

Problem

A circle has center on the side $\displaystyle AB$ of the cyclic quadrilateral $\displaystyle ABCD$ . The other three sides are tangent to the circle. Prove that $\displaystyle AD + BC = AB$ .

Solution

Let the circle have center $\displaystyle O$ and radius $\displaystyle r$ , and let its points of tangency with $\displaystyle BC, CD, DA$ be $\displaystyle E, F, G$ , respectively. Since $\displaystyle OEFC$ is clearly a cyclic quadrilateral, the angle $\displaystyle COE$ is equal to half the angle $\displaystyle GAO$ . Then