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circumcenter of APD is foot of perpendicular from A on BC, NP = NQ, cyclic BPQC

Source: 2015 NZOMC Camp Selections p7

January 10, 2021
geometrycircumcirclecyclic quadrilateralCircumcenter

Problem Statement

Let ABCABC be an acute-angled scalene triangle. Let PP be a point on the extension of ABAB past BB, and QQ a point on the extension of ACAC past CC such that BPQCBPQC is a cyclic quadrilateral. Let NN be the foot of the perpendicular from AA to BCBC. If NP=NQNP = NQ then prove that NN is also the centre of the circumcircle of APQAPQ.
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