Question

$AE$ is the bisector of the exterior $\angle CAD$ meeting $BC$ produced in $E$. If $AB=10cm$, $AC=6cm$ and $BC=12cm$, find $CE$.

Answer 1

In figure,

$AE$ is the bisector of the exterior $\angle CAD$.

$AB=10cm$, $AC=6cm$ and $BC=12cm$

$CE=x$

We know that, The external bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle . [Vertical angle bisector theorem]

In $\Delta ABC$, $AD$ is the bisector of $\angle A$.

$\Rightarrow \frac{BE}{CE}=\frac{AB}{AC}$

$\Rightarrow \frac{12+x}{x}=\frac{10}{6}$

$\Rightarrow 6(12+x)=10x$

$\Rightarrow 72+6x=10x$

$\Rightarrow 72=10x-6x$

$\Rightarrow 72=4x$

$\Rightarrow \frac{72}{4}=x$

$\Rightarrow 18=x$

Therefore, $CE=18cm$