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5. ABC is an isosceles triangle with \small AB=AC. Draw \small AP\perp BC to show that \small \angle B=\angle C.

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Consider \Delta ABP  and  \Delta ACP,

(i)   \angle APB\ =\ \angle APC\ =\ 90^{\circ}       (Since it is given that AP is altitude.)

(ii)   AB\ =\ AC                                      (Isosceles triangle)

(iii)   AP\ =\ AP                                     (Common in both triangles)

Thus by RHS axiom we can conclude that :           

           \Delta ABP\ \cong \Delta ACP

Now, by c.p.c.t.we can say that : 

           \angle B\ =\ \angle C

Posted by

Sanket Gandhi

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