Question

The angle bisector of a triangle divides the opposite side into two parts proportional to the other two sides of the triangle.Prove it.

Answer 1

Step:Proof

Let $ABC$be the triangle.

$AD$is the internal bisector of$\angle BAC$ which meets $BC$ at $D$.

We have to prove $\frac{BD}{DC}=\frac{AB}{AC}$ .

Draw $CE$ parallel to $AD$ meet $BA$produced at $E$

$AC$ is the transversal.

$\angle DAC=\angle ACE-----\left(i\right)$ (alternate interior angle)

$\angle BAD=\angle AEC-----\left(ii\right)$ (corresponding angle)

$\angle BAD=\angle DAC-----\left(iii\right)$ (since $AD$ is the angle bisector of $\angle A$)

From (i), (ii), and (iii), we have

$\angle ACE=\angle AEC$

In $△ACE,AE=AC$

Sides opposite to equal angles are equal.

In $△BCE$,$CE$ parallel to $DA$

$\Rightarrow \frac{BD}{DC}=\frac{AB}{AC}$

Hence proved.