There is another way to identify a probability of $1/4$, by geometry alone.
Consider any three points $A,B,C$ such that no two are on a common diameter. We can pair this arrangement with another, equally probability-dense arrangement by swapping $A$ for the diametrically opposite point $A'$, and we can do the same with $B$, $C$ or any combination of the three points.
Then we have eight equally probability-dense arrangements, four of which are congruent with the other four ($180°$ rotations). Of the four noncongruent arrangements one will have only acute angles when the triangle is drawn and the other three will have an obtuse angle (which would be exchanged for an acute angle, leaving both other angles acute, by swapping the appropriate vertex). Ergo one outbid four arrangements has the center inside the triangle.