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.
Given: In isosceles
To prove: ∠MOC = ∠ABC
Proof:
In ∆ABC,
$\Rightarrow \frac{1}{2} \angle A B C=\frac{1}{2} \angle A C B$
Now, in ∆OBC, ∠MOC is an exterior angle
Hence, ∠MOC = ∠ABC (Given, OB is the bisector of ∠B)
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