Orthologic Triangles in a Quadrilateral


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

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

|Activities| |Contact| |Front page| |Contents| |Geometry|

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.


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

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.

Maxwell Theorem and orthologic triangles

  1. Maxwell's Theorem
  2. Gergonne and Medial Triangles Are Orthologic
  3. Orthologic Triangles in a Quadrilateral
  4. Pedal Triangle and Isogonal Conjugacy
  5. Projective Proof of Maxwell's Theorem
  6. Maxwell's Theorem by Way of Trigonometric Ceva

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

 63808667

Search by google: