Let ABC be a triangle inscribed in the circle (O) and P is a point inside the triangle ABC. Let D be a point on (O) such that AD⊥AP. The line CD cuts the perpendicular bisector of BC at M. The line AD cuts the line passing through B and is perpendicular to BP at Q. Let N be the reflection of Q through M. Prove that CN⊥CP.
Solution
Let K,L be the midpoints of PQ,PN. Since PBQA is cyclic ∠KPB=∠BAQ=∠DCB. Hence the isosceles triangles KPB and MCB are similar. The line BM cuts PQ at R. We have ∠LMC=∠LMB−∠BMC=∠MRQ−∠BKP=∠KBM. Easily seen KB=KP=ML and BM=MC. Hence, △KBM=△LMC. We deduce LC=KM=LP=LN. This, triangle PCN is right at C. We are done. □
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