olymmath

BALKAN-SL 2019 G2

    Let be a triangle △ABC\triangle A B C with m(∠ABC)=75∘m(\angle A B C)=75^{\circ} and m(∠ACB)=45∘m(\angle A C B)=45^{\circ}. The angle bisector of ∠CAB\angle C A B intersects CBC B at the point DD. We consider the point E∈(AB)E \in(A B), such that DE=DCD E=D C. Let PP be the intersection of the lines ADA D and CEC E. Prove that PP is the midpoint of the segment ADA D.

    Dataset text NuminaMath olympiads. Quoted with source.

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