# Projections on Internal and External Angle Bisectors

What Is This About?

A Mathematical Droodle

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

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.

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 | = (180^{o} - ∠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 BII_{C}.)

### References

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

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

63797537 |