Question

Question 9
Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other of 0. BO is produced to a point M. Prove that $\angle MOC=\angle ABC$.

Answer 1

Given lines, OB and OC are the angle bisectors of $\angle B$ and $\angle C$ of an isosceles $\Delta ABC$ such that AB = AC which intersect each other at O and BO is produced to M.

To prove :
$\angle MOC=\angle ABC$.

Proof :
$In\Delta ABCAB=AC$ [given]
$\Rightarrow \angle ACB=\angle ABC$ [angles opposite to equal sides are equal]
$\frac{1}{2}\angle ACB=\frac{1}{2}\angle ABC$ [dividing both sides by 2]
$\Rightarrow \angle OCB=\angle OBC...\left(i\right)$
[since OB and OC are the bisectors of $\angle B$ and $\angle C$]
now $\angle MOC=\angle OBC+\angle OCB$
[exterior angle of a triangle is equal to the sum of two interior angles]
$\Rightarrow \angle MOC=\angle OBC+\angle OBC$ [from Eq (i)]
$\Rightarrow \angle MOC=2\angle OBC$
$\Rightarrow \angle MOC=\angle ABC$ [since, OB is the bisector of $\angle B]$