Probability of triangle to be obtuse 
Two points $A,B$ fixed on a plane(distance = 2).
C - random choosen point inside circle with radius $R$ with center at the center of $AB$
Find probability of triangle $ABC$ to be obtuse

My thoughts: 


*

*If $C$ lies in the circle - $ABC$ will be rectangular

*Opposite the larger angle is large side, so using cosine theorem: 
$$
b^2+c^2<a^2=4R^2,
$$
where $a$ - the diagonal, $b=AC, c=BC$ 


And then I don't know...

Any such triangle $ABC$ with $AB$ diameter and $C$ any point within the circle will be obtuse.
Proof:-

(Note:-$AO=OB=r$)
Let,C be any point in the circle.$AC$ is extended to meet the circumference at $X$.So,$\angle AXB=\alpha=90^0$ and $\angle \beta>0^0$.So,$\angle ACB=\gamma=\alpha+\beta=90$+something $>0.$So,$\angle ACB$ will always be obtuse and thus $\triangle ACB$ is always obtuse.
