1.

1. Prove that the internal bisector of an angle of a triangledivides the opposites side internally is the ratio of theCorresponding sides containing the angle.​

Answer»

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)


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