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Tangency point of mixtilinear incircle is isogonal to...

Source: European Girl's MO 2013, Problem 5

April 11, 2013
geometrycircumcirclegeometric transformationhomothetyincenterEGMOEGMO 2013

Problem Statement

Let Ω\Omega be the circumcircle of the triangle ABCABC. The circle ω\omega is tangent to the sides ACAC and BCBC, and it is internally tangent to the circle Ω\Omega at the point PP. A line parallel to ABAB intersecting the interior of triangle ABCABC is tangent to ω\omega at QQ.
Prove that ∠ACP=∠QCB\angle ACP = \angle QCB.
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