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) (∵ - By eq (1), \space (2) and (3), we get,
\begin{align} &\angle ACE = \angle AEC \\ \implies & AC = AE && (\because \text {sides opposite to equal angles in a triangle are equal}) \end{align} - Now in \triangle BCE, DA \parallel CE,
\begin{align} &\implies \dfrac{ BD }{ DC } = \dfrac{ BA }{ AE } && \text{(by basic proportionality theorem)}\\ & \implies \dfrac{ BD }{ DC } = \dfrac{ AB }{ AC } \end{align} - Hence, the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.