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<BCP < <BHP wanted, parallelogram related to orthocenter

Source: Ukrainian Geometry Olympiad 2020, IX p3, X p2

April 27, 2020
geometryparallelogramorthocenteranglesangle inequalities

Problem Statement

Let HH be the orthocenter of the acute-angled triangle ABCABC. Inside the segment BCBC arbitrary point DD is selected. Let PP be such that ADPHADPH is a parallelogram. Prove that ∠BCP<∠BHP\angle BCP< \angle BHP.
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