Answer :

**ABC is a triangle; AD is the exterior bisector of ∠A and meets BC produced at D; BA is produced to F.**

Given:

Given:

**To prove:**

**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. '

By BPT,

But AC = AE (proved above)

or (proved).

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