$a^2+b^2=2Rc$,where $R$ is the circumradius of the triangle.Then prove that $ABC$ is a right triangle If in a triangle $ABC$,$c$ is the longest side and $a^2+b^2=2Rc$,where $R$ is the circumradius of the triangle.Then prove that $ABC$ is a right triangle.
$a^2+b^2=2Rc\Rightarrow a^2+b^2=\frac{c^2}{\sin C}$
$\sin C=\frac{c^2}{a^2+b^2}$,how to proceed ahead?I am stuck. 


In a $\triangle ABC$,  $c$ is the longest side  and $a^2+b^2=2Rc$,
  where $R$ is the circumradius of the triangle. Prove that $\triangle ABC$ is a right triangle.

We have 
\begin{align}
\sin C&=\frac{c^2}{a^2+b^2},
\\
\cos C&=
\tfrac12 \frac{a^2+b^2-c^2}{ab},
\end{align}
hence
\begin{align}
1&=\left(\frac{c^2}{a^2+b^2}\right)^2
+\left(\tfrac12 \frac{-c^2+a^2+b^2}{ab}\right)^2
\end{align}
$\Rightarrow$
\begin{align}
0&=
\frac{(a^2+b^2-c^2)
((6 a^2 b^2+a^4+b^4) c^2+b^4 a^2+b^2 a^4-a^6-b^6)
}{
4 a^2 b^2 (a^2+b^2)^2
}
\end{align}
So one solution for $c$ in terms of $a$ and $b$
certainly corresponds to a right triangle, $c^2=a^2+b^2$.
Assuming $a>b$, the other one is
\begin{align}
c^2&=
\frac{
(a^2+b^2)(a^2-b^2)^2
}{
(a^2+b^2)^2+4 a^2 b^2
}
\end{align}
or,
\begin{align}
c^2&=a^2
\frac{
\left(1-\tfrac{b^2}{a^2}\right)^2
}{
1+4\tfrac{b^2}{a^2+b^2}+\tfrac{b^2}{a^2}
}
<a^2,
\end{align}
hence for this solution $c$ is not the longest side,
the answer is: it is indeed a right triangle.
Edit
Example solution with "wrong" (that is, not right at all) triangle 
for which $a^2+b^2=2Rc$ ($c<a$):

\begin{align}
a&=2,\ b=1,\ c=\tfrac3{41}\sqrt{205}\approx 1.047645443
\\
\rho&=\tfrac12(a+b+c)=\tfrac32+\tfrac3{82}\sqrt{205}
\\
S&=\sqrt{\rho(\rho-a)(\rho-b)(\rho-c)}=\tfrac9{41}
\\
R&=\frac{abc}{4S}=\tfrac16\sqrt{205}
\end{align}
Check:
\begin{align}
a^2+b^2&=5
\\
2Rc&=2\cdot \tfrac16\sqrt{205}\cdot \tfrac3{41}\sqrt{205}=5.
\end{align}
