Question

in tri ABC the bisector of interior angle A meets BC in D. The bisector of exterior angle A meets bc produced in E. PROVE THAT:- BD:BE=CD:CE. HINT: for the bisector of angle A which is exterior of tri BAC, AB:AC=BE:CE. please give the proof?....................

Answer 1

Solutions 

We know that AD is the interior angle bisector of angle A . 

AE will be the exterior angle bisector of angle. 

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In Δ αβc 

$\frac{AB}{AC}$ = $\frac{BD}{CD}$   ........... (angle bisector theorem ) 

$\frac{AB}{AC}$ = $\frac{BE}{CE}$ ..............( exterior angle bisector theorem )

Now we get 

$\frac{BD}{CD}$ = $\frac{BE}{CE}$ 

$\frac{BD}{CE}$ = $\frac{CD}{CE}$ 

We also know that $\frac{a}{b}$ = $\frac{c}{d}$

= $\frac{a}{c}$ = $\frac{b}{d}$ 

BD : BE  = CD : CE                                       ( Hence proved ) ∴