# Property of Angle Bisectors II

What is this about?

A Mathematical Droodle

What if applet does not run? |

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

Copyright © 1996-2018 Alexander BogomolnyAngle 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.

What if applet does not run? |

(The proof below was suggested by Prof. W. McWorter.)

### Proof

Let E on AD (or its extension) be such that

∠ACB = ∠ABE.

Then triangles ACD and ABE are similar. In particular,

∠AEB = ∠ADC.

From where (and if necessary passing to the supplementary angles)

∠BED = ∠BDE.

Which implies that ΔDBE is isosceles:

(*) | BD = BE. |

On the other hand, similarity of triangles ACD and ABE implies

AB/AC = BE/DC,

which combined with (*) gives the desired proportion. (The proof is different from a more standard one.)

This property of angle bisectors is one way to show that the three 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 Bogomolny64221359 |