A circle has center on the side AB of the cyclic quadrilateral ABCD. The other three sides are tangent to the circle. Prove that AD + BC = AB.
Solution
Let the circle touch AD, CD, BC at L, M, N respectively. Take X on the line AD on the same side of A as D, so that AX = AO, where O is the center of the circle. Now the triangles OLX and OMC are congruent: OL = OM = radius of circle, ∠OLX = ∠OMC = 90o, and ∠OXL = 90o - A/2 = (180o - A)/2 = C/2 (since ABCD is cyclic) = ∠OCM. Hence LX = MC. So OA = AL + MC. Similarly, OB = BN + MD. But MC = CN and MD = DL (tangents have equal length), so AB = OA + OB = AL + LD + CN + NB = AD + BC.
Solutions are also available in Murray S Klamkin, International Mathematical Olympiads 1978-1985, MAA 1986, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
22 Oct 1998