# Proving that the given triangle is isosceles

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

• Have you drawn a diagram? That usually helps when dealing with geometry – lioness99a Sep 27 '18 at 14:15

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.