Question

In Fig., $D$ is a point on side $BC$ of $△ABC$ such that $\frac{BD}{CD}=\frac{AB}{AC}.$ Prove that $AD$ is the bisector of $\angle BAC$.

Answer 1

$D$ is a point on $BC$ of $ABC.$

and $\frac{BD}{CD}=\frac{AB}{AC}$

Let us construct $BA$ to $E$ such that $AE=AC.$ Join $CE.$

Now, as $AE=AC$,

$\frac{BD}{CD}=\frac{AB}{AC}$

$\Rightarrow \frac{BD}{CD}=\frac{AB}{AE}$

Also, $\angle AEC=\angle ACE$ (angles opp. to equal sides of a triangle are equal) ....... (i)

By converse of Basic Proportionality Theorem,

$DA\parallel CE$

$\angle DAC=\angle ACE$ ...... (ii) ...[Alternate angles]

$\angle BAD=\angle AEC$ ......... (iii) ...[Corresponding ∠s]

Also, $\angle AEC=\angle ACE$ ...[From (i)]

and $\angle BAD=\angle DAC$ ...[From (ii) and (iii)]

So, $AD$ is the bisector $\angle BAC$.