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 )

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

AD =  AD                                                              ( Common side )

So,

ADB ADC                                              ( By SAS rule )                  ( Hence proved )

ii ) B =  C                                                     ( By CPCT )                        ( Hence proved )

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

iv ) ADB  =  ADC                                        ( By CPCT )                --- ( 1 )

And

ADB +  ADC  =  180°                                  ( Linear pair angles )

Now we substitute value from equation 1 we get :

ADB +  ADB  =  180° ,

2 ADB =  180° ,

ADB   = 90° , From equation 1 we get :

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

AD BC                    ( Hence proved )


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a)congruence proved by SSS property
   AD=AD(common line)                          s
   AB=AC(given/isoceles triangle)           s
   BD=DC(given)                                      s
b)BD=DC(given/corresponding parts in congruent triangles are equal:CPCT)
​c)AD bisects <A into 2 equal halves
   and BC=180* 
   so <BDA and <ADC =90*(180/2)
therefore AD perpendicular to BC
 
 
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