# 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.

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Copyright © 1996-2018 Alexander Bogomolny## Orthologic Triangles in a Quadrilateral

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.

The solution is due to Darij Grinberg.

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Let S be the orthocenter of ΔACD. Then,

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,

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Copyright © 1996-2018 Alexander Bogomolny