Projections on Internal and External Angle Bisectors
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10 May 2015, Created with GeoGebra


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The applet below provides an illustration to a problem from an outstanding collection by T. Andreescu and R. Gelca (see also [Exercices de Géométrie by F. G.-M.]):

Prove that the four projections of vertex A of ΔABC on the exterior and interior angles bisectors at B and C are collinear.

10 May 2015, Created with GeoGebra

Let M and N denote projections of A onto the interior angle bisectors at B and C, and let P and Q be the projections on the respective exterior angle bisectors. Let us prove that P lies on MN. Since the interior and exterior angles bisectors at a vertex of a triangle are perpendicular, quadrilateral AMBP is a rectangle. Hence

∠AMP = ∠ABP.

We have

∠ABP= (180o - ∠B)/2
 = (∠A + ∠C)/2.

If I is the incenter, quadrilateral ANIM is cyclic for it has two opposite right angles. Therefore,

∠AMN = ∠AIN.

But ∠AIN is exterior to Δ; hence ∠AIN = ∠A/2 + ∠B/2 = ∠AMN. Which implies that points M, N, and P are collinear. That Q lies on the same line is shown in a similar manner.

(Nathan Bowler has observed that P, N and M are collinear since they lie on the Simson line of A with respect to BIIC.)


  1. T. Andreescu, R. Gelca, Mathematical Olympiad Challenges, Birkhäuser, 2004, 5th printing, 1.2.4 (p. 8)
  2. F. G.-M., Exercices de Géométrie, Éditions Jacques Gabay, sixiéme édition, 1991, p. 327

Related material

Angle Bisector

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  • Angle Bisector Theorem
  • All about angle bisectors
  • Angle Bisectors in Ellipse
  • Angle Bisectors in Ellipse II
  • Angle Bisector in Equilateral Trapezoid
  • Angle Bisector in Rectangle
  • Property of Angle Bisectors
  • Property of Angle Bisectors II
  • A Property of Angle Bisectors III
  • External Angle Bisectors
  • Angle Bisectors On Circumcircle
  • Angle Bisectors in a Quadrilateral - Cyclic and Otherwise
  • Problem: Angle Bisectors in a Quadrilateral
  • Triangle From Angle Bisectors
  • Property of Internal Angle Bisector - Hubert Shutrick's PWW
  • Angle Bisectors Cross Circumcircle
  • For Equality Choose Angle Bisector
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