Projections on Internal and External Angle Bisectors
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A Mathematical Droodle
Explanation
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Copyright © 1996-2012 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 | = (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.)
References
- T. Andreescu, R. Gelca, Mathematical Olympiad Challenges, Birkhäuser, 2004, 5th 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|
|Store|
Copyright © 1996-2012 Alexander Bogomolny
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