Thébault's Problem III, Proof
The applet illustrates a synthetic solution for Thébault's Problem III. See if you can surmise what is it about before proceeding to the explanation.  |Activities| |Contact| |Front page| |Contents| |Geometry| |Store| Copyright © 1996-2012 Alexander Bogomolny The applet illustrates a solution to Thébault's Problem III by Jean-Louis Ayme (2003).
By the construction, PE and QG are both perpendicular to BC; so PE||QG. By Sawayama's Lemma, both EF and GH pass through the incenter I of ΔABC. Triangles DHG and QGH being isosceles in D and Q, DQ is
Mutatis mutandi, DP is
As the bisectors of two adjacent and supplementary angles are perpendicular, Michel Cabart found an alternative ending: D is barycenter of E and G, with coefficients DG and DE, thus
Let α be the angle of DP with respect to DE
IE = GE·sinα thus IE = (GE sinα)·(QD/QD) Replacing in (1) yields
thus I is barycenter of Q and P with coefficients sin²α and cos²α and, as such, is collinear with both. References|Activities| |Contact| |Front page| |Contents| |Geometry| |Store| Copyright © 1996-2012 Alexander Bogomolny |
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