Question

Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O. BO is produced to a point M. Prove that ∠MOC = ∠ABC.

Answer 1

Given:
Lines OB and OC are the bisectors of ∠B and ∠C of an isosceles ΔABC such that AB = AC which intersect each other at O and BO is produced to M.

To prove:
∠MOC = ∠ABC

Proof:
In ΔABC,
AB = AC (given)
∠ACB = ∠ABC (angles opposite to equal sides are equal)
$\frac{1}{2}$∠ACB = $\frac{1}{2}$∠ABC (dividing both sides by 2)

Therefore,
∠OCB = ∠OBC …… (1)
(Since, OB and OC are the bisector of ∠B and ∠C)

Now, from equation (1), we have
∠MOC = ∠OBC + ∠OCB
∵ sum of interior angle = Exterior angle

∠MOC = 2∠OBC
⇒∠MOC = 2∠ABC × $\frac{1}{2}$ ​= ∠ABC
(Since, OB is the bisector of ∠B)

Hence, proved.