Question

In $△$ ABC ( Fig.4.59 ), if $\angle$ 1 = $\angle$ 2, prove that $\frac{AB}{AC}=\frac{BD}{DC}$.

Answer 1

Given : $\angle 1=\angle 2$

Applying sine rule to the $△ADB$ and $△ADC$

$\Rightarrow \frac{AB}{\sin \angle BDA}=\frac{BD}{\sin \angle 1}$ ---(1)

$\Rightarrow \frac{AC}{\sin \angle CDA}=\frac{DC}{\sin \angle 2}=\frac{DC}{sin\angle 1}$ ---(2) (As $\angle 1=\angle 2$)

But $\sin \angle CDA=\sin (180-\angle BDA)=\sin \angle BDA$

Applying in (2)

$\frac{AC}{sin\angle BDA}=\frac{DC}{sin\angle 1}$ -------(3)

Dividing equation (1) by equation (3) ,

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