Given a triangle $ABC$ with angles a,b & c, prove that if $\sin^2(a) + \sin^2(b) + \sin^2(c) = 2$ then the triangle is right angled (has an angle of $90^o$).
If I assume the triangle is right angled and have AB, AC and BC as sides, with BC being the base, then I can say that statement is true, since $\sin^2(a)$ would be 1 where $a=90^o$.
Also, since it's a right angled triangle we can say that $\sin^2(b) = AC^2/BC^2$ and $\sin^2(c) = AB^2/BC^2$. The sum of $\sin^2(b) + \sin^2(c)$ would be equal to $(AC^2+AB^2)/BC^2$ which is 1 (pythagoras theorem). Hence the sum of squares of sines of all 3 angles is indeed 2.
But how do I prove this? As in, I'm only given the starting equation, how do I get from there to a right angled triangle?