Question

The bisector of the exterior $\angle CAF$ of a $\Delta ABC$, intersects the side BC produced at D. Show that $\frac{BA}{AC}=\frac{BD}{DC}$.

Answer 1

Given $AD$ is bisector $\angle CAF$

To prove : - $\left(\frac{BA}{AC}\right)=\left(\frac{BD}{DC}\right)$

Prove $\to$ draw $EC\parallel AD$

Hence $\angle ACE=\angle CAD$ (Alternate interior angles)

$\angle AEC=\angle FAD$ (Corresponding angles)

Thus, $\angle AEC=\angle ACE\Rightarrow AE=AC$

Now $EC\parallel AD$

Thus by basic proportional theorem

$\left(\frac{BA}{AE}\right)=\frac{BD}{CD}$

$\Rightarrow \left(\frac{BA}{AC}\right)=\left(\frac{BD}{DC}\right)$ since $AE=AC$

Hence Proved