P and Q are points on the equal sides AB and AC respectively of an isosceles triangle ABC such that AP = CQ. Moreover, neither P nor Q is a vertex of ABC. Prove that the circumcircle of the triangle APQ passes through the circumcenter of the triangle ABC.

Solution by Steve Dinh, a.k.a. Vo Duc Dien

Let the circumcircle of triangle APQ intercept the bisector of ∠A of triangle ABC at I. We have ∠PAI = ∠QAI and PI = QI.

Since AQIP is cyclic, we have ∠AQI + ∠API = 180 or 180 - ∠API = ∠AQI

Now consider the two triangles BPI and AQI. We have BP = AQ, PI = QI, and ∠BPI = 180° - ∠API = ∠AQI; the two triangles are congruent and thus AI = BI.

Since AI is the bisector of ∠BAC and ABC is an isosceles triangle with AB = AC, AI is also the altitude to BC, and BI = CI.

Therefore, BI = CI = AI or I is the circumcenter of ΔABC.

A dynamic illustration of the problem is available elsewhere. Additional five solutions are available on a separate page.