Question

prove that the internal bisector of an angle of a traingle divides the opposite side internally in the ratio of the other two sides.4 mark
answer please it's urgent​

Answer 1

Answer:

Given:

Let

ABC

be the triangle

AD

be internal bisector of

∠BAC

which meet

BC

at

D

To prove:

D C

B D

=

A C

A B

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

⇒

D C

B D

=

A E

B A

....(Thales Theorem)

⇒

D C

B D

=

A C

A B

....

(∴

AE=

AC)

Step-by-step explanation:

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