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  • AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that (i) AD bisects BC

2.(i)  AD is an altitude of an isosceles triangle ABC in which \small AB=AC. Show that

   (i) AD bisects BC

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Consider \Delta ABD   and   \Delta ACD , 

(i)  AB\ =\ AC               (Given)

(ii)  AD\ =\ AD              (Common in both triangles)

(iii)  \angle ADB\ =\ \angle ADC\ =\ 90^{\circ}

Thus by RHS axiom we can conclude that :

    \Delta ABD\ \cong \ \Delta ACD

Hence by c.p.c.t. we can say that  :         BD\ =\ CD      or        AD bisects BC.  

 

Posted by

Sanket Gandhi

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