1. ## Problem involving Similarities!

Given triangle ABC with AB > AC. The bisectors of the interior and exterior angles at A intersect line BC at points D and E, respectively.

PROVE that:

2. Hello, Kaloda!

I haven't solve this yet, but I have some observations.

$\displaystyle \text{Given }\Delta ABC\text{ with }AB > AC.$

$\displaystyle \text{The bisectors of the interior and exterior angles at }A$
. . . $\displaystyle \text{intersect line }BC\text{ at points }D\text{ and }E\text{, respectively.}$

$\displaystyle \text{Prove that: }\:\dfrac{\sqrt{AD^2+AE^2}}{CD} - \dfrac{\sqrt{AD^2+AE^2}}{BD} \:=\: 2$

The bisectors of an interior angle and its exterior angle are perpendicular.

. . Hence: .$\displaystyle \angle DAE = 90^o\:\text{ and }\: \sqrt{AD^2 + AE^2} \:=\:DE$

3. tnx ... that's what i need ...