Question

In the figure, $AB=AC$ and $OB$ and $OC$ are angle bisectors of $\angle B$ and $\angle C.$ Prove that $\angle BOC=\angle ACD$.

Answer 1

Let $\angle ABO=\angle OBC=x$ and $\angle ACO=\angle OCB=y.$

As it is given that $AB=AC.$ So,

$\angle B=\angle C$

$\Rightarrow 2x=2y$

$\Rightarrow x=y\dots \left(i\right)$

In $\Delta ABC$,

$\angle ABC+\angle BCA+\angle CAB={180}^o$

$\Rightarrow 2x+2y+\angle CAB={180}^o$

$\Rightarrow \angle A={180}^o-2x-2y\dots \left(ii\right)$

By using exterior angle property,

$\angle ACD=\angle A+\angle B$

$={180}^o-2x-2y+2x$ [From $\left(ii\right)$]

$={180}^o-2y\dots \left(iii\right)$

Now, by using angle sum property in $△OBC,$

$\angle OBC+\angle OCB+\angle BOC={180}^o$

$\Rightarrow x+y+\angle BOC={180}^o$

$\Rightarrow y+y+\angle BOC={180}^o$ [From $\left(i\right)$]

$\Rightarrow \angle BOC={180}^o-2y\dots \left(iv\right)$

From equations $\left(iii\right)$ and $\left(iv\right)$ we can conclude that,

$\angle BOC=\angle ACD$

Hence, proved.