Find the angles of a triangle given the area For a triangle $ABC$ if:
The area = $\frac{(a+b)^2}{8}$
Find its angles .
My attempt:
$\frac{1}{2} a b \sin{C} = \frac{(a+b)^2}{8}$
$2ab(2\sin{C} -1) = a^2 + b^2 $
$2\sin{C} -1 = \frac{a^2 +b^2}{2ab}$
$2\sin{C} -1$ =$ \frac{a^2 + b^2 -c^2}{2ab} + \frac{c^2}{2ab}$
= $\cos{C} +  \frac{c^2}{2ab}$
But i could not go on any more i need a hint ?

You are on the right track. I have edited out some calculation mistakes including the missing $2$ in front of the $\sin C$. I think it's this mistake that makes the progress hard.
Notice the following inequality:
$$2\sin{C} -1 = \frac{a^2 +b^2}{2ab} \geq1 \implies \sin C \geq 1$$
We can immediately infer that $C$ is $90$ degrees and $a^2+b^2=2ab$ which implies the other two angles  are both $45$ degrees.
