Desargues in the Bride's Chair
(with Pythagoras)
In the famous Bride's Chair diagram that underlies one of Euclid's proofs (I.47) of the Pythagorean proposition, two lines AK and BF serendipitously cross on the altitude from C. The fact remains true for a more general Bride's Chair where the reference triangle is not required to be right. The proof is not difficult: it uses the fact that perpendiculars from A to BF and from B to AK meet on the C-altitude of triangle ABC, thus reducing the proof to the existence of the orthocenter in a triangle.
In case of the right triangle ABC, professor W. McWorter came up with what I believe amounts to a proof without words. A proof that requires no or little explanations. (I apologize for a change in notations.)
The configuration of two squares ACxb and BCya that meet orthogonally at the shared vertex C is symmetric with respect to their common diagonal ab. Which implies concurrency of three lines: AB, ab, and xy at, say, point z. Giving a little thought to the diagram, we may notice the presence of two triangles ABC and abc such that x, y, and z are the crosspoints of the sides, BC and bc, AC and ac, and AB and ab, respectively. The fact of concurrency we have observed says that the two triangles ABC and abc are perspective from a line: the three intersection points x, y, z are collinear. By Desargues' theorem, they are also perspective from a point, which means that the lines Aa, Bb, and Cc are concurrent.
Now, Cc is of course the altitude on which Aa and Bb allegedly meet. To prove this note that the line Cc is a diagonal of rectangle Cycx. The crosspoint of two diagonals of a rectangle serves as their midpoint, so that Cc is a median of triangle Cxy, and hence the altitude of triangle ABC and also of triangle ABc.
Desargues' Theorem
Copyright © 1996-2008 Alexander Bogomolny
|
