MathDB
2 circles and 1 line concurrent, <ABC=90^o, <CPA = < BAC, <BQC = < CBA

Source: Switzerland - Swiss ΜΟ 2016 p5

July 15, 2020
equal anglesconcurrencyconcurrentgeometry

Problem Statement

Let ABCABC be a right triangle with ∠ACB=90o\angle ACB = 90^o and M the center of ABAB. Let GG br any point on the line MCMC and PP a point on the line AGAG, such that ∠CPA=∠BAC\angle CPA = \angle BAC . Further let QQ be a point on the straight line BGBG, such that ∠BQC=∠CBA\angle BQC = \angle CBA . Show that the circles of the triangles AQGAQG and BPGBPG intersect on the segment ABAB.
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