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.
On
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.
$$ \triangle ABC = (r-a).\sqrt{2ra-a^2} + (r-b).\sqrt{2rb-b^2} + (r-c).\sqrt{2rc-c^2}$$
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 $BC$.
$$Area=\frac{AB.BC.CA}{4r}$$