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Harmonic convex quadriteral and angles

Source: tuymaada 2003

March 11, 2006
geometrycircumcircleangle bisectorgeometry unsolved

Problem Statement

In a convex quadrilateral ABCDABCD we have AB⋅CD=BC⋅DAAB\cdot CD=BC\cdot DA and 2∠A+∠C=180∘2\angle A+\angle C=180^\circ. Point PP lies on the circumcircle of triangle ABDABD and is the midpoint of the arc BDBD not containing AA. It is known that the point PP lies inside the quadrilateral ABCDABCD. Prove that ∠BCA=∠DCP\angle BCA=\angle DCP
Proposed by S. Berlov
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