Orthologic Triangles in a Quadrilateral

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The following problem has been posted at the MathLinks forum:

Let ABCD be a convex quadrilateral. The lines parallel to AD and CD through the orthocenter H of ΔABC intersect BC and AB in P and Q, respectively. Prove that the perpendicular through H to PQ passes through the orthocenter of ΔACD.

Solution

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The solution is due to Darij Grinberg.

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Let S be the orthocenter of ΔACD. Then, CS ⊥ AD. On the other hand, HQ||AD. Thus, CS ⊥ HQ, so that the line HQ is the perpendicular to the line CS through the point Q. Similarly, HP ⊥ AS. Finally, since H is the orthocenter of triangle ABC, the line BH is the B-altitude of this triangle, i.e. it is the perpendicular to the line CA through the point B. Thus, the perpendiculars from the vertices B, P, Q of ΔBPQ to the sidelines AC, AS, CS of the triangle ACS concur (namely, these perpendiculars are the lines BH, HP, HQ and thus concur at the point H).

Hence, by Maxwell's theorem, it follows that the perpendiculars from the vertices A, C, S of the triangle ACS to the sidelines PQ, BQ, BP of the triangle BPQ concur. Now, the perpendicular from the point A to the line BP is simply the perpendicular from the point A to the line BC, i.e. the A-altitude of triangle ABC, and similarly, the perpendicular from the point C to the line BQ is the C-altitude of triangle ABC; hence, the point of concurrence, lying on both of these perpendiculars, must be the point of intersection of the A-altitude and the C-altitude of triangle ABC, i.e. simply the orthocenter H of triangle ABC. We thus obtain that the point H lies on the perpendicular from the point S to the line PQ. In other words, SH ⊥ PQ. Thus, the perpendicular to the line PQ through the point H passes through the point S, i.e. through the orthocenter of triangle ACD. And the problem is solved.

1. Gergonne and Medial Triangles Are Orthologic
2. Pedal Triangle and Isogonal Conjugacy
3. Orthologic Triangles in a Quadrilateral

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