Question

In the given figure, $AB=AC$. Then, $\angle BOC=\angle ACD$.
If the above statement is true then mention answer as 1, else mention 0.

Answer 1

Given, $AB=AC$
Hence, $\angle ABC=\angle ACB$ ...(isosceles triangle property)
Thus, $\frac{1}{2}\angle ABC=\frac{1}{2}\angle ACB$

or, $\angle ABO=\angle OBC=\angle ACO=\angle OCB$ ...(OB bisects $\angle B$ and OC bisects $\angle C$)
Now, In $△OBC$,
$\angle OBC+\angle OCB+\angle BOC={180}^o$
$\angle BOC=180-2\left(\angle OCB\right)$
$\angle BOC=180-\angle ACB$
$\angle BOC=\angle ACD$ ...(Since, $\angle ACB+\angle ACD=180$)