Proving that the given triangle is isosceles

Question

I have the following geometry problem with me:

Let $O$ and $O_1$ be the centres of the incircle and excircle opposite to $A$ of triangle $ABC$. The Perpendicular bisector of $OO_1$ meets lines $AB$ and $AC$ at $L$ and $N$ respectively. Given that the circumcircle of triangle $ABC$ touches $LN$, prove that $ABC$ is isosceles

I have proceeded a little in the following way:

Let $OO_1$ intersect $LN$ at $E$, it is clear that $E$ is the midpoint of $LN$, also $OO_1$ is perpendicular to $LN$. Also let $D$ be the point of tangency of the circumcircle and the line $LN$, which implies that if $O_2$ is the circumcentre of $ABC$, then $O_2D$ is perpendicular to $LN$, which implies that $OO_1$ is parallel to $O_2D$. If $F$ is the foot of perpendicular from $E$ onto $BC$, then $BF=BC$ and $EF$ passes through $O_2$. I know at this point that if $OO_1$ coincides with $O_2D$ then the problem is done, alternatively I can prove that $BC$ is parallel to $LN$. However I am unable to proceed from here.

Note: From power of the point, I also know that $LB*AL = LD^2$ and $NC*NA = ND^2$, and $AL = AN$ and I feel that this needs to be used to prove the above statement of mine. ANy help from here please

Answer 1

Try to prove the following lemma by angle chasing or analytic geometry:

In any triangle the circumcircle passes through the midpoint of the segment which unites the incenter with an excenter.

Answer 2

Note that if LN is tangent to the circumcircle it can intersect(touch) it only at a single point. So $E$ and $D$ are the same point. So $A$, $O_2$ and $O$ are collinear. Simple angle chasing does the rest of the job.