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Length condition and prove angles sum to 90

Source: China TST Test 2 Day 2 Q5

March 11, 2019
geometryChina TST

Problem Statement

Let MM be the midpoint of BCBC of triangle ABCABC. The circle with diameter BCBC, ω\omega, meets AB,ACAB,AC at D,ED,E respectively. PP lies inside △ABC\triangle ABC such that ∠PBA=∠PAC,∠PCA=∠PAB\angle PBA=\angle PAC, \angle PCA=\angle PAB, and 2PM⋅DE=BC22PM\cdot DE=BC^2. Point XX lies outside ω\omega such that XM∥APXM\parallel AP, and XBXC=ABAC\frac{XB}{XC}=\frac{AB}{AC}. Prove that ∠BXC+∠BAC=90∘\angle BXC +\angle BAC=90^{\circ}.
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