IMO 1997 SL 20

Origin: IRE

Problem

Let D be an internal point on the side BC of a triangle ABC. The line AD meets the circumcircle of ABC again at X. Let P and Q be the feet of the perpendiculars from X to AB and AC, respectively, and let \gamma be the circle with diameter XD. Prove that the line PQ is tangent to \gamma if and only if AB = AC.

Solution

To avoid dividing into cases regarding the position of the point X, we use oriented angles. Let R be the foot of the perpendicular from X to BC. It is well known that the points P, Q, R lie on the corresponding Simson line. This line is a tangent to \gamma (i.e., the circle XDR) if and only if \anglePRD = \angleRXD. We have \anglePRD = \anglePXB = 90◦−\angleXBA = 90◦−\angleXBC + \angleABC = 90◦−\angleDAC + \angleABC and \angleRXD = 90◦−\angleADB = 90◦+ \angleBCA −\angleDAC; hence \anglePRD = \angleRXD if and only if \angleABC = \angleBCA, i.e, AB = AC.