If ∆ABC is isosceles with AB = AC and C (0, r) is the incircle of the ∆ABC touching BC at L. Prove that L bisects BC.
If ∆ABC is isosceles with AB = AC and C (0, r) is the incircle of the ∆ABC touching BC at L. Prove that L bisects BC.

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 $BC$.