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


Answer:


Step by Step Explanation:
  1. We are given a ABC and AD is the angle bisector of A.
    Now, construct CEDA as given in the figure below such that it meets BA produced at E.
    A C B D E
  2. Since CEDA,
    DAC=ACE.....(1) (Alternate interior Angles)BAD=AEC.....(2) (Corresponding Angles)Also, BAD=DAC.....(3) (
  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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