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Geometry Difficulty 8.3 Shortlist Prove it Hong Kong

Two circles Γ\Gamma and Ω\Omega intersect at two distinct points AA and BB. Let PP be a point on Γ\Gamma. The tangent at PP to Γ\Gamma meets Ω\Omega at the points CC and DD, where DD lies between PP and CC, and ABCDABCD is a convex quadrilateral. The lines CACA and CBCB meet Γ\Gamma again at EE and FF respectively. The lines DADA and DBDB meet Γ\Gamma again at SS and TT respectively. Suppose the points P,E,S,F,B,T,AP, E, S, F, B, T, A lie on Γ\Gamma in this order. Prove that PC,ET,SFPC, ET, SF are parallel.

Solution

Since
TEC=TEA=TBA=DBA=DCA=DCE, \angle TEC = \angle TEA = \angle TBA = \angle DBA = \angle DCA = \angle DCE,
we get PC//ETPC//ET. Similarly, since
FSD=FSA=CBA=PDA=PDS, \angle FSD = \angle FSA = \angle CBA = \angle PDA = \angle PDS,
we get PC//SFPC//SF.

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