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Prove CPM=ACB

Source: Japan TST 2016 P1

January 25, 2021
geometryangle bisectorcircumcircle

Problem Statement

In acute triangle △ABC\triangle ABC, MM is the midpoint of BCBC. Let ℓ\ell be the angle bisector of ∠BAC\angle BAC. The tangent at CC to the circumcircle of △ABC\triangle ABC meets ℓ\ell at DD. Point HH lies on ℓ\ell and satisfies ∠ABH=90∘\angle ABH=90^{\circ}. The circumcircles of △CDH\triangle CDH and △ABC\triangle ABC intersect again at PP. Prove that ∠CPM=∠ACB\angle CPM=\angle ACB.
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