Problem involving Similarities!

**Kaloda** · November 23rd 2010, 04:32 AM

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:
((sqrt (AD^2+AE^2)) / CD) - (sqrt (AD^2+AE^2)) / BD = 2

**Soroban** · November 26th 2010, 03:44 AM

Hello, Kaloda!

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

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

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

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

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

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

**Kaloda** · November 26th 2010, 04:07 AM

here's the pic ...
http://yfrog.com/evhonorsproblem380p

**Kaloda** · November 26th 2010, 04:50 AM

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

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