Question

In figure D is a point on side BC of $\Delta ABC$ such that $\dfrac{{BD}}{{CD}} = \dfrac{{AB}}{{AC}}$. Prove that AD is a bisector of $\angle BAC$.

Answer 1

Hint: As we are given that ratio of sides are equal that is $\dfrac{{BD}}{{CD}} = \dfrac{{AB}}{{AC}}$, hence for any triangle we can use sine formula that is $\dfrac{{\sin a}}{a} = \dfrac{{\sin b}}{b} = \dfrac{{\sin c}}{c}$ .
Apply this on both the triangles ABD and ADC.

Complete step-by-step answer:
We are given $\Delta ABC$ in which AD is produced on BC such that $\dfrac{{BD}}{{CD}} = \dfrac{{AB}}{{AC}}$

We are given that $\dfrac{{BD}}{{CD}} = \dfrac{{AB}}{{AC}}$
As we know in any triangle sine formula is applicable that is
$\dfrac{{\sin a}}{a} = \dfrac{{\sin b}}{b} = \dfrac{{\sin c}}{c}$
So in $\Delta ABD$ using sine formula
$\dfrac{{\sin \angle BAD}}{{BD}} = \dfrac{{\sin \angle ADB}}{{AB}}\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \left( 1 \right)$
In $\Delta ADC$
$\dfrac{{\sin \angle ADC}}{{AC}} = \dfrac{{\sin \angle DAC}}{{DC}}\,\,\,\,\,\,\, \to \left( 2 \right)$
Here we know that $\angle ADB = 180 - \angle ADC$
Taking sine on both sides
$\sin \angle ADB = \sin \left( {180 - \angle ADC} \right)$
And we know $\sin \left( {180 - \theta } \right) = \sin \theta $
So $\sin \angle ADB = \sin \angle ADC$
So in equation (1) we get
$
  \dfrac{{\sin \angle BAD}}{{BD}} = \dfrac{{\sin \angle ADC}}{{AB}}\,\,\, \\
  \sin \angle ADC = \dfrac{{AB}}{{BD}}\sin \angle BAD\,\,\,\,\,\,\,\,\,\, \to (3) \\
 $
Putting this in equation (2) we get
$
  \dfrac{{AB}}{{BD.AC}}\sin \angle BAD\, = \dfrac{{\sin \angle DAC}}{{DC}} \\
  \dfrac{{AB}}{{AC}}\sin \angle BAD\, = \dfrac{{BD}}{{CD}}\sin \angle DAC \\
 $
We also know that $\dfrac{{BD}}{{CD}} = \dfrac{{AB}}{{AC}}$
As given in question:
So here we get
$\sin \angle BAD = \sin \angle DAC$
Hence we can say
$\angle BAD = \angle DAC$
Hence it is proved that AD bisects angle A.

Note: This is the angle bisector theorem that in triangle ABC if AD bisects angle A then sides are proportional that means that $\dfrac{{BD}}{{CD}} = \dfrac{{AB}}{{AC}}$ or vice versa.