Question

The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the corresponding sides containing the angle. Prove it.

Answer 1

Given:
Let $ABC$ be the triangle
$AD$ be internal bisector of $\angle BAC$ which meet $BC$ at $D$
To prove: $\frac{BD}{DC}=\frac{AB}{AC}$
Draw $CE\parallel DA$ to meet $BA$ produced at $E$
Since $CE\parallel DA$ and $AC$ is the transversal.
$\angle DAC=\angle ACE$ (alternate angle ) .... (1)
$\angle BAD=\angle AEC$ (corresponding angle) .... (2)
Since $AD$ is the angle bisector of $\angle A$
$\therefore \angle BAD=\angle DAC$ .... (3)
From (1), (2) and (3), we have
$\angle ACE=\angle AEC$
In $△ACE$,
$\Rightarrow AE=AC$
($\therefore$ Sides opposite to equal angles are equal)
In $△BCE$,
$\Rightarrow CE\parallel DA$
$\Rightarrow \frac{BD}{DC}=\frac{BA}{AE}$ ....(Thales Theorem)
$\Rightarrow \frac{BD}{DC}=\frac{AB}{AC}$ ....$(\therefore AE=AC)$