Problem:
Let â–³ABC be an acute triangle, with O as its circumcenter. Point H is the foot of the perpendicular from A to line BC, and points P and Q are the feet of the perpendiculars from H to the lines AB and AC, respectively.
Given that
AH2=2â‹…AO2
prove that the points O,P, and Q are collinear.
Solution:
First, since AP⋅AB=AH2=AQ⋅AC, it follows that PQCB is cyclic. Consequently, we have AO⊥PQ. Let K be the foot of A onto PQ, and let D be the point diametrically opposite A. Thus A,K,O,D are collinear.
Since quadrilateral KQCD is cyclic (∠QKD=∠QCD=90∘), we have
