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:
- We are given a △ABC and AD is the angle bisector of ∠A.
Now, construct CE∥DA as given in the figure below such that it meets BA produced at E. - Since CE∥DA,
∠DAC=∠ACE.....(1) (Alternate interior Angles)∠BAD=∠AEC.....(2) (Corresponding Angles)Also, ∠BAD=∠DAC.....(3) (∵AD is bisector of ∠A) - By eq (1), (2) and (3), we get,
∠ACE=∠AEC⟹AC=AE(∵sides opposite to equal angles in a triangle are equal) - Now in △BCE,DA∥CE,
⟹BDDC=BAAE(by basic proportionality theorem)⟹BDDC=ABAC - Hence, the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.