# In the adjoining figure, ∆ABC is an isosceles triangle in which AB = AC and AD is the bisector of ∠A. Prove that: (i) ∆ADB ≡ ∆ADC (ii) ∠B = ∠C (iii) BD = DC (iv) AD ⊥ BC

Dear Student,

Please find below the solution to the asked query:

i ) In $∆$ ADB and $∆$ ADC

AB =  AC                                                              (  Given )

$\angle$ BAD =  $\angle$ CAD                                               ( As given AD is angle bisector of $\angle$ A )

AD =  AD                                                              ( Common side )

So,

$∆$ ADB $\cong$ $∆$ ADC                                              ( By SAS rule )                  ( Hence proved )

ii ) $\angle$ B =  $\angle$ C                                                     ( By CPCT )                        ( Hence proved )

iii )  BD  =  CD                                                     ( By CPCT )                        ( Hence proved )

iv ) $\angle$ ADB  =  $\angle$ ADC                                        ( By CPCT )                --- ( 1 )

And

$\angle$ ADB +  $\angle$ ADC  =  180$°$                                  ( Linear pair angles )

Now we substitute value from equation 1 we get :

$\angle$ ADB +  $\angle$ ADB  =  180$°$ ,

2 $\angle$ ADB =  180$°$ ,

$\angle$ ADB   = 90$°$ , From equation 1 we get :

$\angle$ ADB  = $\angle$ ADC  =  90$°$  , From this relation we can say that :

AD $\perp$ BC                    ( Hence proved )

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a)congruence proved by SSS property