Q5 of 23 Page 1

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 -


Construction: Draw CEAD 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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