The Mirror Property of Altitudes
What is this about?
A Mathematical Droodle
Explanation
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Copyright © 1996-2012 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.
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.
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).
References
- R. Honsberger, The Butterfly Problem and Other Delicacies from the Noble Art of Euclidean Geometry II, TYCMJ, 14 (1983), pp. 154-158.
- C. W. Trigg, Mathematical Quickies, Dover, 1985
|Activities|
|Contact|
|Front page|
|Contents|
|Geometry|
|Store|
Copyright © 1996-2012 Alexander Bogomolny
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