MathDB

2021.G5

Part of Cono Sur Shortlist - geometry

Problems(1)

<CBQ = <PBA wanted, feet of altitudes and circumcircle related

Source: 2021 Cono Sur Shortlist G5 https://artofproblemsolving.com/community/c1088686_cono_sur_shortlist__geometry

10/31/2022
Let △ABC\vartriangle ABC be a triangle with circumcenter OO, orthocenter HH, and circumcircle ω\omega. AA′AA', BB′BB' and CC′CC' are altitudes of △ABC\vartriangle ABC with A′A' in BCBC, B′B' in ACAC and C′C' in ABAB. PP is a point on the segment AA′AA'. The perpenicular line to B′C′B'C' from PP intersects BCBC at KK. AA′AA' intersects ω\omega at M≠AM \ne A. The lines MKMK and AOAO intersect at QQ. Prove that ∠CBQ=∠PBA\angle CBQ = \angle PBA.
geometryequal angles