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KQ=QH and fixed circle wanted, altitudes, 2 common tangents to circles

Source: 2017 XX All-Ukrainian Tournament of Young Mathematicians named after M. Y. Yadrenko, Qualifying p3

May 6, 2021
fixed circlefixedgeometryUkrainian TYM

Problem Statement

The altitude AH,BTAH, BT, and CRCR are drawn in the non isosceles triangle ABCABC. On the side BCBC mark the point PP; points XX and YY are projections of PP on ABAB and ACAC. Two common external tangents to the circumscribed circles of triangles XBHXBH and HCYHCY intersect at point QQ. The lines RTRT and BCBC intersect at point KK. a). Prove that the point QQ lies on a fixed line independent of choiceP P. b). Prove that KQ=QHKQ = QH.