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The Mirror Property of Altitudes: What is this about?
A Mathematical Droodle


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Explanation

Copyright © 1996-2008 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.

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.


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 http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


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Proof

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.

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

  AF = AE·BH / BE.

Similarly,

  AG = AD·CH / CD.

By Ceva's theorem we have

  AE/BE · BH/CH · CD/AD = 1,

or,

  (BH·AE/BE)·(CD/(CH·AD)) = 1,

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

References

  1. R. Honsberger, The Butterfly Problem and Other Delicacies from the Noble Art of Euclidean Geometry II, TYCMJ, 14 (1983), pp. 154-158.

Copyright © 1996-2008 Alexander Bogomolny

28774360Page copy protected against web site content infringement by Copyscape


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