The exterior bisector of the angle A of triangle ABC meets BC on D tha

1.	The exterior bisector of the angle A of triangle ABC meets BC on D than prove that AB/AC=BD/BC
Answer» 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).

The exterior bisector of the angle A of triangle ABC meets BC on D than prove that AB/AC=BD/BC

Answer»

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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The exterior bisector of the angle A of triangle ABC meets BC on D than prove that AB/AC=BD/BC

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