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Question

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=ABC2 .....2

Similarly, OCB=C2 .....3

In ∆BOC,

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

ABC2+C2+BOC=180° Using 2 and 3ABC+C2+BOC=180°2ABC2+BOC=180° Using 1ABC+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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