Let ABCD be a convex quadrilateral with the line CD tangent to the circle on diameter AB. Prove that the line AB is tangent to the circle on diameter CD if and only if BC and AD are parallel.
Solution
If AB and CD are parallel, then AB is tangent to the circle on diameter CD if and only if AB = CD and hence if and only if ABCD is a parallelogram. So the result is true.
Suppose then that AB and DC meet at O. Let M be the midpoint of AB and N the midpoint of CD. Let S be the foot of the perpendicular from N to AB, and T the foot of the perpendicular fromM to CD. We are given that MT = MA. OMT, ONS are similar, so OM/MT = ON/NS and hence OB/OA = (ON - NS)/(ON + NS). So AB is tangent to the circle on diameter CD if and only if OB/OA = OC/OD which is the condition for BC to be parallel to AD.
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
19 Oct 1998