IMO 1984 SL 14

Origin: ROM

Problem

Let ABCD be a convex quadrilateral for which the circle with diameter AB is tangent to the line CD. Show that the circle with diameter CD is tangent to the line AB if and only if the lines BC and AD are parallel.

Solution

Let M and N be the midpoints of AB and CD, and let M ′, N ′ be their projections on CD and AB, respectively. We know that MM ′ = AB/, and hence SABCD = SAMD +SBMC +SCMD = 1 2(SABD +SABC)+ 1 4AB \cdotCD. (1) The line AB is tangent to the circle with diameter CD if and only if NN ′ = CD/2, or equivalently, SABCD = SAND + SBNC + SANB = 1 2(SBCD + SACD) + 1 4AB \cdot CD. By (1), this is further equivalent to SABC + SABD = SBCD + SACD. But since SABC + SACD = SABD + SBCD = SABCD, this reduces to SABC = SBCD, i.e., to BC \parallelAD.