The Mirror Property of Altitudes
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A Mathematical Droodle

19 February 2015, Created with GeoGebra


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Copyright © 1996-2018 Alexander Bogomolny

The applet may suggest the following assertion:

Let P be a point on the altitude AH of triangle ABC such that H is located between B and C. Extend BP and CP to their intersection with AC and AB in D and E, respectively. Then AH is the angle bisector of ∠DHE, or equivalently ∠BHE = ∠CHD.

mirror property of altitudes, problem

This is known as the Mirror Property of the altitudes and hence of the orthic triangle. (See also the mirror property in disguise proven using Pascal's theorem.) It was the main device in solving Fagnano's problem.


The proof is by Professor Hiroshi Haruki.

Draw a line through A parallel to BC, and let F and G denote the intersections with the line of HE and HD extended. The assertion will follow from the fact that AF = AG.

mirror property of altitudes, solution

By construction, triangles BEH and AEF are similar: AF/BH = AE/BE, giving

AF = AE·BH / BE.


AG = AD·CH / CD.

By Ceva's theorem we have

AE/BE · BH/CH · CD/AD = 1,


(BH·AE/BE)·(CD/(CH·AD)) = 1,

which reduces to AF/AG = 1, yielding AF = AG.

This problem was posted by Nathan Altshiller Court in the Mathematics Magazine (37, November 1964, p. 338). The solution had an extra line parallel to the base pass through point P, instead of A. The problem had been reproduced in C. W. Trigg's collection (#135).


  1. R. Honsberger, The Butterfly Problem and Other Delicacies from the Noble Art of Euclidean Geometry II, TYCMJ, 14 (1983), pp. 154-158.
  2. C. W. Trigg, Mathematical Quickies, Dover, 1985

Related material

  • The Mirror Property of Altitudes via Pascal's Hexagram
  • The Angle Bisectors
  • Optical Property of Ellipse
  • Fagnano's Problem
  • Parabolic Mirror, Illustration
  • Existence of the Orthocenter
  • All about altitudes
  • |Activities| |Contact| |Front page| |Contents| |Geometry|

    Copyright © 1996-2018 Alexander Bogomolny