Question

In the given figure, $AD$ is the bisector of the exterior $\angle A$ of $\Delta ABC$. Seg $AD$ intersects the side $BC$ produced in $D$. Prove that $\frac{BD}{CD}=\frac{AB}{AC}$.

Answer 1

Since $\angle CAK$ is the exterior angle of $△ABC$, we can say that $m\angle CAK=\angle B+\angle C$ (Exterior Angle Theorem).

Therefore, $m\angle CAD=\frac{\angle B+\angle C}{2}$ (since $AD$ is an angle bisector)

$m\angle BAD=\angle A+\frac{\angle B+\angle C}{2}=\frac{2\angle A+\angle B+\angle C}{2}=\frac{\angle A+180}{2}=\frac{\angle A}{2}+90$

$m\angle ADB=180-(\angle B+\angle BAD)=180-(\angle B+\angle A+\frac{\angle B+\angle C}{2})=180-\angle A-\frac{3\angle B+\angle C}{2}$

Now, in $△ABD$, applying Sine rule,

$\frac{BD}{\sin \left(\angle BAD\right)}=\frac{AB}{\sin \left(\angle ADB\right)}$

$\frac{BD}{\sin (90+\frac{\angle A}{2})}=\frac{AB}{\sin \left(\angle ADB\right)}$

$\frac{BD}{AB}=\frac{\cos \left(\frac{\angle A}{2}\right)}{\sin \left(\angle ADB\right)}$ .....$\left(1\right)$

In $△ACD$, applying Sine rule,

$\frac{CD}{\sin \left(\angle CAD\right)}=\frac{AC}{\sin \left(\angle ADB\right)}$

$\frac{CD}{\sin \left(\frac{\angle B+\angle C}{2}\right)}=\frac{AC}{\sin \left(\angle ADB\right)}$

$\frac{CD}{AC}=\frac{\sin \left(\frac{180-\angle A}{2}\right)}{\sin \left(\angle ADB\right)}=\frac{\cos \left(\frac{\angle A}{2}\right)}{\sin \left(\angle ADB\right)}$ .....$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, we get

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

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

Hence proved.