Answer :

Given - Δ ABC is a triangle Ad is exterior bisector of ∠A and meets BC at D

To prove -

Construction: Draw CE∥AD to meet AB at E

Proof -

⇒ In Δ ABC, CE ∥ AD cut by AC

∴ ∠CAD = ∠ACE - - - alternate angles

In ΔABC CE

⇒ In Δ ABC, CE ∥ AD cut by AB

∴ ∠FAD = ∠AEC - - - corresponding angles

∴∠FAD = ∠CAD - - given

∴∠ACE = ∠AEC

AC = AE—isosceles triangle

⇒ In Δ BAD, CE ∥ DA

∴AE = DC

∴AB = BD

But AC = AE

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