MathDB

1

Part of 2016 Japan TST

Problems(1)

Prove CPM=ACB

Source: Japan TST 2016 P1

1/25/2021
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.
geometryangle bisectorcircumcircle