Find possible values of angle C In an acute triangle $ABC$, $O$ is the circumcenter, $H$ is the orthocenter and $G$ is the
centroid. Let $OD$ be perpendicular to $BC$ and $HE$ be perpendicular to $CA$, with $D$ on $BC$ and $E$
on $CA$. Let $F$ be the midpoint of $AB$. Suppose the areas of triangles $ODC, HEA$ and $GFB$ are
equal. Find all the possible values of angle $C$.
My approach :
Let $R$ be the circumradius of $△ABC$ and $∆$ its area. We have $OD = R \cos A$ and
$DC =\frac{a}{2}$, so
$$[ODC] = \frac{1}{2}· OD · DC$$ $$=\frac{1}{2}· R \cos A · R \sin A $$ $$=\frac{1}{2}. R^2.\sin A \cos A .$$
Again $HE = 2R \cos C \cos A$ and $EA = c \cos A$. Hence
$$[HEA] = \frac{1}{2}· HE · EA$$ 
$$=\frac{1}{2}· 2R \cos C \cos A · c \cos A$$
 $$=2R^2.\sin C .\cos C.{\cos}^2A .$$
Further
$$[GFB] = \frac{∆}{6}=\frac{1}{6}· 2R^2.\sin A.\sin B.\sin C$$ $$=\frac{1}{3}.R^2.\sin A.\sin B.\sin C$$
What to do next?  Any help would be greatly appreciated.

From $[ODC]=[HEA]$ one gets
$$
\tag{1}
\sin C\cos C={1\over4}{\sin A\over \cos A}.
$$
From $[ODC]=[GFB]$, taking into account that $\sin B=\sin(A+C)=\sin A\cos C+\cos A\sin C$, one gets
$$
\sin A \sin C\cos C+\cos A \sin^2C={3\over2}\cos A,
$$
that is, using $(1)$:
$$
{1\over4}{\sin^2 A\over \cos A}+\cos A \sin^2C={3\over2}\cos A,
$$
whence:
$$
\tag{2}
\sin^2 C={7\cos^2 A-1\over4\cos^2A}.
$$
Squaring $(1)$ and inserting there $(2)$ one finally gets
$$
20\cos^4A-9\cos A+1=0,
\quad\hbox{that is:}\quad
\cos^2A={1\over4}
\ \ \hbox{or}\ \ 
\cos^2A={1\over5}.
$$
From $(2)$ one then obtains
$$
\sin^2 C={3\over4}\ \hbox{and}\ C=60°,
\ \ \hbox{or}\ \ 
\sin^2C={1\over2}\ \hbox{and}\ C=45°.
$$
The first case obviously leads to an equilateral triangle, while the second case is not trivial.
