The lengths of the sides of a triangle are $m^{2}+n^{2}$,$2mn$,and $m^{2}-n^{2}$,where $m > n > 0$. Prove that the triangle is a right-angled triangle.

(N/A) Let the sides of the triangle be $a = m^{2}-n^{2}$,$b = 2mn$,and $c = m^{2}+n^{2}$.
To prove that the triangle is a right-angled triangle,we need to check if the square of the longest side is equal to the sum of the squares of the other two sides (Pythagoras Theorem).
Here,$c = m^{2}+n^{2}$ is the longest side since $m > n > 0$.
Calculate $a^{2} + b^{2}$:
$a^{2} + b^{2} = (m^{2}-n^{2})^{2} + (2mn)^{2}$
$= (m^{4} - 2m^{2}n^{2} + n^{4}) + 4m^{2}n^{2}$
$= m^{4} + 2m^{2}n^{2} + n^{4}$
$= (m^{2}+n^{2})^{2}$
$= c^{2}$
Since $a^{2} + b^{2} = c^{2}$,the triangle satisfies the Pythagoras Theorem.
Therefore,the triangle is a right-angled triangle.