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Geometry from Iran TST 2017

Source: 2017 Iran TST third exam day1 p2

April 26, 2017
geometryTST

Problem Statement

Let PP be a point in the interior of quadrilateral ABCDABCD such that: ∠BPC=2∠BAC  ,  ∠PCA=∠PAD  ,  ∠PDA=∠PAC\angle BPC=2\angle BAC \ \ ,\ \ \angle PCA = \angle PAD \ \ ,\ \ \angle PDA=\angle PAC Prove that: ∠PBD=∣∠BCA−∠PCA∣\angle PBD= \left | \angle BCA - \angle PCA \right |
Proposed by Ali Zamani
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