The bisectors of ∠B and ∠C of an isosceles triangle with AB = AC intersect each other at a point O.


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 ">ΔABC, AB = ACOB and OC are bisectors of B and C, respectively.

To prove: MOC = ABC


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)


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