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 proceding to the explanation.

Copyright © 1996-2008 Alexander Bogomolny

The applet illustrates a solution to Thébault's Problem III by Jean-Louis Ayme (2003).

By the construction, PE and QG are bothe 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

1. the perpendicular bisector of GH,
2. the D-internal bisector of ΔDHG.

Mutatis mutandi, DP is

1. the perpendicular bisector of EF,
2. 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 at 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.

References

1. J.-L. Ayme, Sawayama and Thébault's Theorem, Forum Geometricorum, v 3 (2003), 225-229

1. Thébault's Problem I
2. Thébault's Problem II
3. Thébault's Problem III
   • Y. Sawayama's Lemma
   • Y. Sawayama's Theorem
   • Thébault's Problem III, Proof (J.-L. Ayme)

Copyright © 1996-2008 Alexander Bogomolny