Question

In triangle ABC, the side AB is greater than the side AC. The bisector of angle a meet BC at D. Prove that BD > DC. SOLVE PLS!!

Answer 1

Answer:

Given: ABC is a triangle; AD is the exterior bisector of 4A and meets BC

produced at D; BA is produced to F.

To prove: AB/AC = BD/DC

Construction: Draw CE||DA to meet AB at E.

Proof: In A ABC. CE||AD cut by AC.

∠CAD = ∠ACE (Alternate angles)

Similarly CE || AD cut by AB

∠FAD = ∠AEC (corresponding angles)

AC = AE (by isosceles △ theorem)

Now in △BAD, CE || DA

AE = DC (BPT)

AB BD

But AC = AE (proved above)

AC = DC

AB = BD or

AB/AC = BD/DC (proved).

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