Question

In the figure, given below, $AB=AC$.
Hence, $\angle BAC=\angle ACD$.

• A. True
• B. False

Answer 1

The correct option is B False

Given, $AB=AC$.
Hence, $\angle ABC=\angle ACB$ (Isosceles triangle property).
Also, $OB$ bisects $\angle B$ and $OC$ bisects $\angle C$.
Thus, $\angle OBC=\angle ABO=\frac{1}{2}\left(\angle B\right)$
and $\angle OCB=\angle ACO=\frac{1}{2}\left(\angle C\right)$.

Since, $\angle B=\angle C$,

hence, $\angle OBC=\angle ABO=\angle ACO=\angle OCB$... $\left(i\right)$.
Now, in $△OBC$,
$\angle OBC+\angle OCB+\angle BOC={180}^o$ (Angles of a triangle)..$\left(I\right)$.

Also, $\angle ACB+\angle ACD={180}^o$ (Angles on a straight line) ...$\left(II\right)$.

Equating $\left(I\right)$ and $\left(II\right)$,
$\angle OBC+\angle OCB+\angle BOC=\angle ACB+\angle ACD$
$\angle OBC+\angle OCB+\angle BOC=\angle ACO+\angle OCB+\angle ACD$.

Hence,$\angle BOC=\angle ACD$ (From $\left(i\right)$).

Therefore, the statement is false and option $B$ is correct.