Thébault's Problem III, Proof
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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.
Copyright © 1996-2009 Alexander Bogomolny
The applet illustrates a solution to Thébault's Problem III by Jean-Louis Ayme (2003).
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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
- the perpendicular bisector of GH,
- the D-internal bisector of ΔDHG.
Mutatis mutandi, DP is
- the perpendicular bisector of EF,
- the D-internal bisector of ΔDEF.
As the bisectors of two adjacent and supplementary angles are perpendicular, DQ ⊥ DP. Therefore, GH||DP and DQ||EF. In the hexagon PEIGQD the sides are pairwise parallel, i.e. concur Point at infinityat infinity and therefore the points of incidence all lie on a line at infinity. By the converse of Pappus' Theorem, the points P, I, Q are collinear.
Michel Cabart found an alternative ending:
D is barycenter of E and G, with coefficients DG and DE, thus
| (1) |
ID = (DG/GE)·IE + (DE/GE)·IG
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Let α be the angle of DP with respect to DE
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IE = GE·sinα thus IE = (GE sinα)·(QD/QD)
IG = GE·cosα thus IG = (GE cosα)·(PD/PD)
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Replacing in (1) yields
| | ID | = (DG sinα/QD)·QD + (DE cosα/PD)·PD |
| | | = (sin²α)·QD + (cos²α)·PD. |
thus I is barycenter of Q and P with coefficients sin²α and cos²α and, as such, is collinear with both.
References
- J.-L. Ayme, Sawayama and Thébault's Theorem, Forum Geometricorum, v 3 (2003), 225-229
- Thébault's Problem I
- Thébault's Problem II
- Thébault's Problem III
- Circles Tangent to Circumcircle
Copyright © 1996-2009 Alexander Bogomolny
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