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Part of 2017 Ukrainian Geometry Olympiad

Problems(1)

angle chasing with 2 midpoints, equal angles given and wanted

Source: Ukrainian Geometry Olympiad 2017, IX p1, X p1, XI p1

12/11/2018
In the triangle ABCABC, A1{{A}_{1}} and C1{{C}_{1}} are the midpoints of sides BCBC and ABAB respectively. Point PP lies inside the triangle. Let ∠BPC1=∠PCA\angle BP {{C}_{1}} = \angle PCA. Prove that ∠BPA1=∠PAC\angle BP {{A}_{1}} = \angle PAC .
geometrymidpointsequal anglesAngle Chasing