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

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 \angleMOC = \angleABC.

Answers (1)

Given: ABC is a isosceles triangle AB = AC

BO is produced to point M.

To prove : \angleMOC = \angleABC

Proof :

AB = AC

\angleABC = \angleACB          (Angles opposite to two equal sides of triangle are equal)

In \triangleOBC,

     \angle BOC +\frac{\angle B}{2}+\frac{\angle C}{2} = 180^{\circ}         (\because OB and OC are bisector of ÐB, ÐC)

\Rightarrow \angle BOC + 2 \times \left ( \frac{\angle B}{2} \right )=180^{\circ}    \left ( \because \angle B = \angle C \right )

\Rightarrow \angle BOC + \angle B = 180^{\circ}\\ \Rightarrow \angle BOC = 180^{\circ}- \angle B \cdots \cdots (i)\\ So\; \angle MOC + \angle BOC = 180^{\circ} \; \; \; (Linear\; pair)\\ \Rightarrow \angle MOC = 180 ^{\circ}-\angle BOC\\ \Rightarrow \angle MOC = 180^{\circ} - (180^{\circ} -\angle B) \; \; (From i)\\ \Rightarrow \angle MOC = 180^{\circ} - 180^{\circ} + \angle B\\ \Rightarrow \angle MOC = \angle B\\ \Rightarrow \angle MOC=\angle ABC

Hence proved

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