In ∆ABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB ≅ seg AC then prove that ED || BC.

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Solution

Given:
ray BD bisects ∠ABC
ray CE bisects ∠ACB.
seg AB ≅ seg AC

In △ABC, ∠ABD = ∠DBC
$\frac{AD}{DC} = \frac{AB}{BC} . . . (I) (By angle bisector theorem)$
In △ABC, ∠BCE = ∠ACE
$\frac{AE}{EB} = \frac{AC}{BC} . . . (II) (By angle bisector theorem)$
From (I) and (II)
$\frac{AD}{DC} = \frac{AE}{EB} (∵ seg AB ≅ seg AC)$
∴ ED || BC (Converse of basic proportional theorem)

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