Question

adjoining figure ad is bisector of an exterior angle of triangle ABC seg ad intersect the side BC produced in ad prove that BD/CD=AB/AC

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Answer 1

Answer:

Step-by-step explanation:

Refer the photo for answer

Answer 2

Apply Basic Proportionality Theorem

Step-by-step explanation:

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

produced at D; BA is produced to F.

To prove: $\frac {BD}{CD} = \frac {AB}{AC}$

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

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

$\angle CAD = \angle ACE$ (Alternate angles)

Similarly CE || AD cut by AB

$\angle FAD = \angle AEC$ (corresponding angles)

AC = AE (by isosceles △ theorem)

Now in △BAD, CE || DA  

AE = DC (Basic Proportionality Theorem)

AB = BD

But AC = AE (proved above)

AC = DC  

AB = BD or

$\frac {AB}{AC} = \frac {BD}{DC}$ (proved)