Question

1. Prove that the internal bisector of an angle of a triangle
divides the opposites side internally is the ratio of the
Corresponding sides containing the angle.​

Answer 1

Answer:

Given:

Let ABC be the triangle

AD be internal bisector of ∠BAC which meet BC at D

To prove:

DC BD= AC AB

Draw CE∥DA to meet BA produced at E

Since CE∥DA and AC is the transversal.

∠DAC=∠ACE (alternate angle ) .... (1)

∠BAD=∠AEC (corresponding angle) .... (2)

Since AD is the angle bisector of ∠A

∴∠BAD=∠DAC .... (3)

From (1), (2) and (3), we have

∠ACE=∠AEC

In △ACE,

⇒AE=AC

(∴ Sides opposite to equal angles are equal)

In △BCE,

⇒CE∥DA

⇒ DC BD = AE BA ....(Thales Theorem)

⇒ DC BD = AC AB ....(∴AE=AC)

Answer 2

Answer:

this is a angle bisector theorem

Step-by-step explanation:

in triangle ABC ,AD is the internal bisector

AB/AC=BD/DC

draw a line through C parallel to AB extend AD to meet line through C at E