Question: The bisectors of ∠B and ∠C of an isosceles triangle with AB = AC intersect each other at a point O. BO is produced to meet AC at a point M. Prove that ∠MOC = ∠ABC.
Solution:
Given: In isosceles ∆">Δ∆ABC, AB = AC; OB and OC are bisectors of ∠B and ∠C, respectively.
To prove: ∠MOC = ∠ABC
Proof:
In ∆ABC,
∵">∵∵AB = AC (Given)
∴">∴∴∠ABC = ∠ACB (Angles opposite to equal sides are equal)
$\Rightarrow \frac{1}{2} \angle A B C=\frac{1}{2} \angle A C B$
⇒">⇒⇒∠OBC = ∠OCB (Given, OB and OC are the bisectors of ∠B and ∠C, respectively) .....(i)
Now, in ∆OBC, ∠MOC is an exterior angle
⇒">⇒⇒∠MOC = ∠OBC + ∠OCB (An exterior angle is equal to the sum of two opposite interior angles)
⇒">⇒⇒∠MOC = ∠OBC + ∠OBC [From (i)]
⇒">⇒⇒∠MOC = 2∠OBC
Hence, ∠MOC = ∠ABC (Given, OB is the bisector of ∠B)
