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$AP + 2PB = CP$.

Source: 2016 239 J5

October 11, 2020
geometry

Problem Statement

Triangle ABCABC in which AB<BCAB <BC, is inscribed in a circle ω\omega and circumscribed about a circle γ\gamma with center II. The line ℓ\ell parallel to ACAC, touches the circle γ\gamma and intersects the arcs BACBAC and BCABCA at points PP and QQ, respectively. It is known that PQ=2BIPQ = 2BI. Prove that AP+2PB=CPAP + 2PB = CP.
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