Question

The equal sides AB and AC of an isosceles triangle ABC are produced to D and E respectively. The
bisectors of the angles CBD and BCE meet at O. Prove that triangleOBC has two equal sides.​

Answer 1

Given :-

• In Δ ABC , AB = AC  
• AB and AC are produced to D & E
• Bisector of ∠ CBD & ∠ BCE  meet at O

To prove :-

• Δ OBC has two equal sides

i.e, OB = OC

Proof :-

∵ AB = AC  

∴ ∠ ABC = ∠ ACB   ....eqn(1)  

(angles opposite to equal sides)

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∵ BO is the bisector of ∠ CBD

∴ ∠ CBD = 2 ∠ CBO  .....eqn(2)

and

∵ CO is the bisector of ∠ BCE

∴ ∠ BCE = 2 ∠ BCO  ......eqn(3)

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∵ ABD is line

∴ ∠ ABC + ∠ CBD = 180° (linear pair)

also,

∵ ACE is a line

∴ ∠ ACB + ∠ BCE = 180° (linear pair)

So, we can conclude that

→∠ ABC + ∠ CBD = ∠ ACB + ∠ BCE

Using eqn (1)

→∠ ABC + ∠ CBD = ∠ ACB + ∠ ABC

→ ∠ CBD = ∠ ACB

Using eqn (2) and (3)

→ 2 ∠ CBO = 2 ∠ BCO

→ ∠ CBO = ∠ BCO

Using angles opp. to equal sides are equal

→  BO = CO   (proved.)

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