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Turkish Mathematical Olympiad

Geometry Difficulty 5.9 AIME, harder Prove it Turkey

The points AA and BB lie on a circle with diameter CDCD and on different sides of the line CDCD. A circle Γ\Gamma passing through the points CC and DD intersects the line segment ACAC at a point EE different from its endpoints, and the line BCBC at a point FF. PP is the point of intersection of the tangent line to Γ\Gamma at EE and the line BCBC, and QQ is a point different from EE lying on the circumcircle of the triangle CEPCEP and satisfying QP=EPQP = EP. SS is the midpoint of the line segment EQEQ and RR is the point of intersection of the lines ABAB and EFEF. Show that the lines DRDR and PSPS are parallel.

Solution

As ABAB is the Simson line for the point DD and the triangle FCEFCE, DRDR is perpendicular to EFEF. We have ∠QEP=∠EQP=∠ECF=∠XEF\angle QEP = \angle EQP = \angle ECF = \angle XEF, where XX is a point on the ray PEPE beyond EE. Therefore QQ, EE, FF are collinear. As PSPS is perpendicular to EQEQ, the result follows.

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