Q. 15.0( 2 Votes )
The bisector of the exterior angle ∠A of Δ ABC intersects side BC produced at D. Prove that .
Given: ABC is a triangle; AD is the exterior bisector of ∠A and meets BC produced at D; BA is produced to F.
Construction: Draw CE||DA to meet AB at E.
Proof: In Δ ABC, CE || AD cut by AC.
∠CAD = ∠ACE (Alternate angles)
Similarly CE || AD cut by AB
∠FAD = ∠AEC (corresponding angles)
Since ∠FAD = ∠CAD (given)
∴ ∠ACE = ∠AEC
AC = AE ( by isosceles Δ theorem)
Now in Δ BAD, CE || DA
(Basic proportionality theorem):If a line is drawn parallel to one side of a triangle intersecting other two sides,
then it divides the two sides in the same ratio. '
But AC = AE (proved above)
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