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.

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Copyright © 1996-2018 Alexander Bogomolny

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

By the construction, PG and QE are both perpendicular to BC; so PG||QC. 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 in a point 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.

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

Let α be the angle of DP with respect to DE

IE = GE·sinα  thus IE = (GE sinα)·(QD/QD)
IG = GE·cosα  thus IG = (GE cosα)·(PD/PD)

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.

A third prove has been discovered by Sohail Farhangi; it is based on a property of right trapezoids proved elsewhere. The right trapezoid to consider is EPQG

a proof of Thebault's third theorem by Sohail Farhangi

For any choice of D, DP⊥DQ because they are bisectors of supplementary angles. Then the property of right trapezoids forces IG and IE to be perpendicular, with I lying on PQ.


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

Related material

Thébault's Problems

  • Thébault's Problem I
  • Thébault's Problem II
  • Thébault's Problem III
  • Y. Sawayama's Lemma
  • Jack D'Aurizio Proof of Sawayama's Lemma
  • Y. Sawayama's Theorem
  • Thébault's Problem III, Proof (J.-L. Ayme)
  • Circles Tangent to Circumcircle
  • Thébault's Problem IV
  • A Property of Right Trapezoids
  • A Lemma on the Road to Sawayama
  • Excircles Variant of Thébault's Problem III
  • In the Spirit of Thebault I
  • Dao's Variant of Thebault's First Problem
  • |Activities| |Contact| |Front page| |Contents| |Geometry|

    Copyright © 1996-2018 Alexander Bogomolny