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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.

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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 $BC$.

$$Area=\frac{AB.BC.CA}{4r}$$

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As AB, BC and AC are chords of the same circle, the lines from the center of the circle to their mid-points will be perpendiculars to the chord. So,

$$ OF \perp AB, OD \perp BC \space and \space OE \perp AC $$ $$ AB = 2.AF = 2. \sqrt{OA^2-OF^2} = 2.\sqrt{OA^2-(ON-FN)^2}$$ $$ AB = 2.\sqrt{r^2-(r-c)^2} = 2.\sqrt{2rc-c^2}$$ $$ \triangle AOB = \frac{1}{2}.OF.AB=\frac{1}{2}.(r-c).2.\sqrt{2rc-c^2}=(r-c).\sqrt{2rc-c^2}$$ We can find the area of triangle BOC and AOC similarly.

$$ \triangle ABC = (r-a).\sqrt{2ra-a^2} + (r-b).\sqrt{2rb-b^2} + (r-c).\sqrt{2rc-c^2}$$

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