<s>[INST] $a^2+b^2=2Rc$,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. 
 [/INST] You are correct to use the sine law.
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \tag{1}$$
If you apply the sine law to sides $a,b$ the main equation becomes:
$$\begin{align}
&4R^2 \sin^2A + 4R^2 \sin^2B = 4R^2 \sin C \implies \\
&\sin^2A + \sin^2B = \sin C\tag{2}
\end{align}$$
For a triangle $C = 180^\circ - (A+B)$, so then
$$\sin C = \sin(A+B)$$
Substituting this into (2), we get:
$$\sin^2A + \sin^2B = \sin (A+B) \tag{3}$$
Can you show that a solution is $A+B=90^\circ$, so that $C=90^\circ$?
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