Question

If $△ABC$ is isosceles with $AB=AC$ and $C(0,r)$ is the incircle of the $△ABC$ touching $BC$ at $L$, prove that $L$ bisect $BC$ ?

Answer 1

$△ABC$ is an Isoceles triangle $C(o,r)$ is the incircle of $△ABC$ touching $BC$ at $I,C(o,r)$ is the incircle of $△ABC$

$△O$ is the point of intersection of angle bisector of $△ABC$

i.e $OB$ bisects $\angle B$ and $OC$ bisects $\angle C$

In $△ABC$

$AB=AC$

$\therefore \angle C=\angle B$ (Equal sides have equal angles opposite to them)

$\Rightarrow \frac{1}{2}<C=\frac{1}{2}<B$

$\Rightarrow \angle OCI=\angle OBI$ ( $OB$ bisects $\angle B$ nad $OC$ bisects $\angle C$)

In $△OBI=\angle OCI$

$\angle OIB=\angle OIC$

$\angle OBI=\angle OCI$

$\therefore OBI\cong △OCI$

$\Rightarrow BI=IC$

Thus $I$ bisect the side $BC$