Question

Prove that
The internal bisector of an angle of a triangle divides the opposite side internally in the
ratio of the 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)

Step-by-step explanation:

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Answer 2

Answer:

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