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altitude, median, bisector, and trisector

Source: Indonesia IMO 2010 TST, Stage 1, Test 1, Problem 4

November 12, 2009
trigonometrygeometrygeometric transformationreflectioncircumcircleangle bisectorsimilar triangles

Problem Statement

Let ABC ABC be a non-obtuse triangle with CH CH and CM CM are the altitude and median, respectively. The angle bisector of ∠BAC \angle BAC intersects CH CH and CM CM at P P and Q Q, respectively. Assume that \angle ABP\equal{}\angle PBQ\equal{}\angle QBC, (a) prove that ABC ABC is a right-angled triangle, and (b) calculate BPCH \dfrac{BP}{CH}. Soewono, Bandung
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