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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?....................



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

AE will be the exterior angle bisector of angle. 


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 ) 
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