# External Angle Bisectors

What is this about?

A Mathematical Droodle

What if applet does not run? |

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

Copyright © 1996-2018 Alexander BogomolnyInternal angle bisectors divide the opposite side in the ratio of the adjacent sides. More accurately,

If, in ΔABC, AD is an angle bisector of angle A, then

AB/AC = DB/DC

Perhaps curiously, the same is true of the external angle bisectors, i.e., bisectors of the external angles. And the proof is in fact the same.

What if applet does not run? |

### Proof

Assume the straight line through C parallel to AD meets AB in E. Then, first of all, ΔAEC is isosceles:

∠AEC = ∠BAD = 90° - ∠BAC/2,

such that

∠ACE = 180° - ∠CAE - ∠AEC = 90° = ∠BAC/2

Therefore, AE = AC, and the required proportion follows from the similarity of triangles BEC and BAD.

This property of angle bisectors is one way to show that one internal and two external angle bisectors in a triangle meet in a point. The result is an immediate consequence of Ceva's theorem.

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

Copyright © 1996-2018 Alexander Bogomolny68969859