Solve Question:
In the given figure, AD is the bisector of the exterior ∠A of Δ ABC . Seg AD intersects the side BC produced in D. Prove that: 


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