OC631 from Crux Mathematicorum Vol. 49, No. 5

OC631 (from Crux Mathematicorum Vol. 49, No. 5)

The point \(O\) is the circumcenter of the triangle \(ABC\), and \(AH\) is its altitude. The point \(P\) is the foot of the perpendicular dropped from the point \(A\) to the line \(CO\). Prove that the line \(HP\) passes through the midpoint of the segment \(AB\).

Proof

Extend \(CP\) to intersect the circumcircle at point \(E\). Since \(A,B,C,E\) are concyclic, we have \(\angle ABH = \angle AEC\). Since \(CE\) is the diameter of the circumcircle and \(AP \perp CE\), we have \(\angle AEC = \angle PAC\). Also, observing that \(A,P,H,C\) are concyclic, we have $ PAC = PHB$. Therefore, by angle chasing, we obtain \[\angle ABH = \angle AEC = \angle PAC = \angle PHB. \] It is worth noting that the above angle chasing holds regardless of \(H\) is inside the triangle \(ABC\) or not.

Let \(Q\) be the intersection point between \(AB\) and \(HP\). Then, we have $ QBH = ABH = PHB = QHB $, Noticing that \(\triangle AHB\) is a right triangle with \(\angle AHB\) being the right angle and \(Q\) lies on \(AB\). It follows that \(\angle QAH = \angle QHA\) and \(QB = QH = QA\). Therefore we conclude that \(Q\) is the midpoint of the segment \(AB\).