In a triangle ABC, AB=AC and BE and CF are bisectors of angle B and angle C respectively. Prove that triangle EBC is congruent to triangle FCB (No figure has been provided)

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Please find below the solution to the asked query:

We form our diagram from given information , As :

Here , AB  =  AC so from base angle theorem we get :

$\angle$ ABC =  $\angle$ ACB  ,                               --- ( 1 )

$\angle$ FBC =  $\angle$ ECB                                  --- ( 2 )                ( As $\angle$ ABC  =  $\angle$ FBC  and $\angle$ ACB  = $\angle$FCB  same angles )

And given BE and CF are angle bisectors of $\angle$ B and $\angle$ C respectively , So

$\angle$ ABE  =  $\angle$ EBC = $\frac{1}{2}$ $\angle$ ABC  and  $\angle$ ACF  =  $\angle$ FCB = $\frac{1}{2}$ $\angle$ ACB , So from equation 1 we get  : 

$\angle$ EBC =  $\angle$ FCB                                  --- ( 3 )

Now in $∆$ EBC and $∆$ FCB 

$\angle$ ECB=  $\angle$ FBC                                                    ( From equation 2 ) 

BC  =  BC                                                                ( Common side )

$\angle$ EBC =  $\angle$ FCB                                                    ( From equation 3 ) 

So,

$∆$ EBC $\cong$ $∆$ FCB                                                      ( By ASA )                ( Hence proved )


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