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