Find the area of the triangle knowing the length of perp bisector from side to circumcircle

Let $$\triangle ABC$$ inscribed in circle with center $$O$$ and radius $$r$$. Let $$D,E,F$$ be the mid-point of side $$BC,CA,AB,$$ respectively then $$OD,OE,OF$$ meet the circumcircle at $$L,M,N$$ respectively. If $$DL=a, EM=b, FN=c$$ then find the area of the triangle in term of $$r,a,b,c$$

Could someone help me with this? I approach using Pythagorean to find radius but end up with everything just equal to each other as it should be. Notice that $$OD=r-a$$ and $$OB=r$$. Hence, by Pythagoras Theorem, you get $$BD=\sqrt{2ar-a^2}=\frac{BC}{2}.$$ Similarly, you can do it for $$AB$$ and $$BC$$.
$$Area=\frac{AB.BC.CA}{4r}$$
$$OF \perp AB, OD \perp BC \space and \space OE \perp AC$$ $$AB = 2.AF = 2. \sqrt{OA^2-OF^2} = 2.\sqrt{OA^2-(ON-FN)^2}$$ $$AB = 2.\sqrt{r^2-(r-c)^2} = 2.\sqrt{2rc-c^2}$$ $$\triangle AOB = \frac{1}{2}.OF.AB=\frac{1}{2}.(r-c).2.\sqrt{2rc-c^2}=(r-c).\sqrt{2rc-c^2}$$ We can find the area of triangle BOC and AOC similarly.
$$\triangle ABC = (r-a).\sqrt{2ra-a^2} + (r-b).\sqrt{2rb-b^2} + (r-c).\sqrt{2rc-c^2}$$