Bisectors of angles B and C of an isosceles triangle ABC with AB = AC intersect each other at O. Show that external angle adjacent to ∠ABC is equal to ∠BOC.

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Solution

∠ABD is the external angle adjacent to ∠ABC.

∆ABC is an isosceles triangle.

AB = AC (Given)

∴ ∠C = ∠ABC .....(1) (In a triangle, equal sides have equal angles opposite to them)

Also, OB and OC are the bisectors of ∠B and ∠C, respectively.

$∴ ∠ OBC = \frac{∠ ABC}{2} . . . . . (2)$

Similarly, $∠ OCB = \frac{∠ C}{2} . . . . . (3)$

In ∆BOC,

∠OBC + ∠OCB + ∠BOC = 180º (Angle sum property of triangle)

$\Rightarrow \frac{∠ ABC}{2} + \frac{∠ C}{2} + ∠ BOC = 180 ° [Using (2) and (3)] \Rightarrow \frac{∠ ABC + ∠ C}{2} + ∠ BOC = 180 ° \Rightarrow \frac{2 ∠ ABC}{2} + ∠ BOC = 180 ° [Using (1)] \Rightarrow ∠ ABC + ∠ BOC = 180 ° . . . . . (4)$

Now,

∠ABD + ∠ABC = 180º .....(5) (Linear pair)

From (4) and (5), we have

∠ABD + ∠ABC = ∠ABC + ∠BOC

⇒ ∠ABD = ∠BOC

Thus, the external angle adjacent to ∠ABC is equal to ∠BOC.

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