Question

In an isosceles triangle with $AB=AC$, the bisectors of $\angle B$ and $\angle C$ meet at $O$. Produce $BO$ upto $M$ then prove that $\angle MOC=\angle ABC$.

Answer 1

In $\Delta ABC$ with $AB=AC$
$\angle B=\angle C$ (given)
In $\Delta BOC$
$\angle MOC=\angle OBC+\angle OCB$
$=\frac{1}{2}(2\angle OBC+2\angle OCB)$
$=\frac{1}{2}(\angle B+\angle C)$
$\because \angle B=\angle C$
$\angle MOC=\frac{1}{2}[\angle B+\angle B]$
$=\frac{1}{2}\ast 2\angle B=\angle ABC$
Hence Proved.