olymmath

BALTIC-WAY 2017 Problem 14

    Let PP be a point inside the acute angle ∠BAC\angle B A C. Suppose that ∠ABP=∠ACP=90∘\angle A B P=\angle A C P=90^{\circ}. The points DD and EE are on the segments BAB A and CAC A, respectively, such that BD=BPB D=B P and CP=CEC P=C E. The points FF and GG are on the segments ACA C and ABA B, respectively, such that DFD F is perpendicular to ABA B and EGE G is perpendicular to ACA C. Show that PF=PGP F=P G.

    Dataset text NuminaMath olympiads. Quoted with source.

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