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.