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 angle BCE meet at O. prove that triangle OBC has two equal sides.​

Answer 1

Given  :  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 angle BCE meet at O

To find:   prove that triangle OBC has two equal sides.​

Solution:

AB = AC

=> ∠B = ∠C

∠CBD = ∠A + ∠B  ( Exterior angle of triangle  = Sum of opposite two interior angles)

∠BCE  = ∠A + ∠C  ( Exterior angle of triangle  = Sum of opposite two interior angles)

∠B  = ∠C

Adding   ∠A on Both sides

=> ∠A + ∠B = ∠A + ∠C

=> ∠CBD = ∠BCE

the bisectors of the angles CBD and angle BCE meet at O

=> ∠CBO = (1/2) ∠CBD  

&  ∠BCO = (1/2) ∠BCE

∠CBD = ∠BCE

=> (1/2) ∠CBD   = (1/2) ∠BCE

=>  ∠CBO =   ∠BCO

=> OB = OC

QED

Hence Proved

triangle OBC has two equal sides.​

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Answer 2

Answer:

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