Question

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 1

Let the triangle be ABC with AD as the bisector of $\angle A$ which meets BC at D.
We have to prove:
$\frac{BD}{DC}=\frac{AB}{AC}$



Draw CE $\parallel$ DA, meeting BA produced at E.
CE $\parallel$DA
Therefore,
$$\begin{gathered} \angle 2=\angle 3(\mathrm{Alternate}\mathrm{angles}) \\ \mathrm{and}\angle 1=\angle 4(\mathrm{Corresponding}\mathrm{angles}) \\ \mathrm{But}, \\ \angle 1=\angle 2 \\ \mathrm{Therefore}, \\ \angle 3=\angle 4 \\ \Rightarrow AE=AC \\ \mathrm{In}△BCE,DA\hspace{0.17em}\parallel CE. \\ \mathrm{Applying}\mathrm{Thales}\text{'}\mathrm{theorem},\mathrm{we}\mathrm{gave}: \\ \frac{BD}{DC}=\frac{AB}{AE} \\ \Rightarrow \frac{BD}{DC}=\frac{AB}{AC} \end{gathered}$$
This completes the proof.