# IN THE FIGURE AB = AC. PROVE THAT BD = BC

Dear Student,

Given :  AB  =  AC , So from base angle theorem we get

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

And
BD  =  BC , So from base angle theorem we get

$\angle$ BDC  =  $\angle$ BCD                                           --- ( 2 )  , $\angle$ ACB  =  $\angle$ BCD ( Same angles ) , So from equation 1 a nd w we get

$\angle$ ABC  =  $\angle$ ACB  = $\angle$ BDC                                     --- ( 3 )

From angle sum property in triangle we get in triangle ABC :

$\angle$ BAC +  $\angle$ ABC + $\angle$ ACB =  180$°$ , Substitute values and get

40$°$ + $\angle$ ABC  +  $\angle$ ABC = 180$°$

2 $\angle$ ABC = 140$°$

$\angle$ ABC =  70$°$ , So From equation 3 we get

$\angle$ ABC  =  $\angle$ ACB  = $\angle$ BDC = 70$°$

And

$\angle$ CBD  = $\angle$ ABC - $\angle$ ABD  = 70$°$ - 30$°$  =  40$°$

And

$\angle$ ADB  = 180$°$ - $\angle$ ABD - $\angle$ BAD  = 180$°$ - 30$°$ - 40$°$  =  110$°$

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Regards

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