Given: In ∆ABC, AB = AC and O is the centre of the circle and radius (r) touches the side BC of ∆ABC at L.
Given to prove : BC’s mid-point is L.
Proof :
$AM$ and $AN$ are the tangents.
So,AN$AM=AN$
But $AB=AC$ (given)
$AB-AN=AC-AM$
$\Rightarrow BN=CM$
Now $BL$and$BN$ are the tangents of the circle from $B$
So, $BL=BN$
Similarly,$CL$ and $CM$ are tangents
$CL=CM$
But $BN=CM$ (proved above)
So, $BL=CL$
Therefore, $L$ is mid-point of $BC$.