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In △EDC and △EBA, we have

∠1=∠2 [Alternate angles]

∠3=∠4 [Alternate angles] and,

∠CED=∠AEB [Vertically opposite angles]

∴ △EDC∼△EBA

⇒

EB

ED

=

EA

EC

⇒

EC

ED

=

EA

EB

.............(I)

It is given that △AED∼△BEC

∴

EC

ED

=

EB

EA

=

BC

AD

From (i) and (ii), we get

EA

EB

=

EB

EA

⇒ (EB)

2

=(EA)

2

⇒ EB=EA

Substituting EB=EA in (ii), we get

EA

EA

=

BC

AD

⇒

BC

AD

=1

⇒ AD=BC [Hence proved]