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Geometry Difficulty 5.2 AIME, harder Prove it Saudi Arabia

Let ABCABC be a triangle inscribed in the circle (O)(O) and PP is a point inside the triangle ABCABC. Let DD be a point on (O)(O) such that ADAPAD \perp AP. The line CDCD cuts the perpendicular bisector of BCBC at MM. The line ADAD cuts the line passing through BB and is perpendicular to BPBP at QQ. Let NN be the reflection of QQ through MM. Prove that CNCPCN \perp CP.

Solution

Figure 1
Let K,LK, L be the midpoints of PQ,PNPQ, PN.
Since PBQAPBQA is cyclic
KPB=BAQ=DCB. \angle KPB = \angle BAQ = \angle DCB.
Hence the isosceles triangles KPBKPB and MCBMCB are similar. The line BMBM cuts PQPQ at RR. We have
LMC=LMBBMC=MRQBKP=KBM. \angle LMC = \angle LMB - \angle BMC = \angle MRQ - \angle BKP = \angle KBM.
Easily seen KB=KP=MLKB = KP = ML and BM=MCBM = MC.
Hence, KBM=LMC\triangle KBM = \triangle LMC. We deduce LC=KM=LP=LNLC = KM = LP = LN.
This, triangle PCNPCN is right at CC. We are done. \square

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