Question

in triangle ABC, BE and CD are the angle bisectors of angle ABC and angle ACB. given that BE=CD. Prove that side AB =AC ​

Answer 1

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Extend the line BC to E

BD and CD are angular bisectors,

∴∠ABD=∠DBC=x and ∠ACD=∠DCE=y

∠ABC=2x and ∠ACE=2y

Consider △ABC,

∠ACE=∠ABC+∠BAC ------exterior angle is equal to sum of interior opposite angle

2y=2x+∠A

y−x=∠A/2 ----(i)

Consider △BCD,

∠DCE=∠DBC+∠BDC ------exterior angle is equal to sum of interior opposite angle

y=x+∠D

y−x=∠D------(ii)

From(i) and (ii)

∠D=1/2 ∠A

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Answer 2

Extend the line BC to E

BD and CD are angular bisectors,

∴∠ABD=∠DBC=x and ∠ACD=∠DCE=y

∠ABC=2x and ∠ACE=2y

Consider △ABC,

∠ACE=∠ABC+∠BAC ------exterior angle is equal to sum of interior opposite angle

2y=2x+∠A

y−x=∠A/2 ----(i)

Consider △BCD,

∠DCE=∠DBC+∠BDC ------exterior angle is equal to sum of interior opposite angle

y=x+∠D

y−x=∠D------(ii)

From(i) and (ii)

∠D=1/2 ∠A