Question

10. In the adjoining figure, AD is the angle bisector of exterior 2 A of A ABC.
BD bisects 2 B. If AB = AC, then prove that
(i) AD || BC
(ii) AB = AD​

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).