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


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Answer:


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^@.
    A C B D 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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