Let ABC be a triangle where ∣AB∣=∣AC∣. Points P and Q are different from the vertices of the triangle and lie on the sides AB and AC, respectively. Prove that the circumcircle of the triangle APQ passes through the circumcenter of ABC if and only if ∣AP∣=∣CQ∣.
Solution
Without loss of generality, we can assume that ∣AP∣≤∣AQ∣. Let O be the circumcenter of ABC. Let R be the intersection point of the bisector of ∠BAC with the circumcircle of the triangle PAQ — we then have ∣RB∣=∣RC∣ (Fig. 21).
Also, ∠APR=180∘−∠AQR=∠CQR and ∣RP∣=∣RQ∣ (since ∠RAP=∠RAQ). So, ∣AP∣=∣CQ∣⟺△APR≅△CQR⟺∣RA∣=∣RC∣⟺R=O (where ∣RA∣=∣RC∣⇒△APR≅△CQR by two sides and obtuse angle).
Fig. 21
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Source: MathNet,
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