### 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. A C B D

Step by Step Explanation:
1. We are given a $\triangle ABC$ and $AD$ is the angle bisector of $\angle A$.
Now, construct $CE \parallel DA$ as given in the figure below such that it meets $BA$ produced at $E$.
2. Since $CE \parallel DA$,
\begin{align} & \angle DAC = \angle ACE && .....(1) \text{ (Alternate interior Angles)} \\ & \angle BAD = \angle AEC && .....(2) \text{ (Corresponding Angles)} \\ & \text{Also, } \angle BAD = \angle DAC && .....(3) \space (\because AD \text{ is bisector of } \angle A) \end{align}
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}
4. 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}
5. Hence, the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.

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