Problem with sine in a right triangle 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?

Notice, in $\Delta ABC$ , we have $a+b+c=180^o$
Given that 
$$\sin^2 a+\sin^2 b+\sin^2 c=2$$ $$\implies \sin^2 a+\sin^2 b+\sin^2 (180^o-(a+b))=2$$ $$\implies \sin^2 a+\sin^2 b+\sin^2 (a+b)=2$$ 
$$\implies (\sin a\cos b+\cos a\sin b)^2=2-\sin^2 a-\sin^2 b$$ $$\implies \sin^2 a\cos^2 b+\cos^2 a\sin^2 b+2\sin a\sin b\cos a\cos b=\cos^2 a+\cos^2 b$$ 
 $$\implies  -(1-\sin^2 a)\cos^2 b-(1-\sin^2 b)\cos^2 a+2\sin a\sin b\cos a\cos b=0$$  $$\implies  -\cos^2 a\cos^2 b-\cos^2 a\cos^2 b+2\sin a\sin b\cos a\cos b=0$$
$$\implies -2\cos a\cos b(\cos a\cos b-\sin a\sin b)=0$$  $$\implies \cos a\cos b\cos(a+b)=0$$ $$ \color{blue}{\text{if } \quad \cos a=0 \implies a=90^o}$$ $$ \color{blue}{\text{if } \quad \cos b=0 \implies b=90^o}$$ $$ \color{blue}{\text{if } \quad \cos (a+b)=0 \implies a+b=90^o \iff c=90^o}$$
We find that above three cases show that $\color{blue}{\Delta ABC}$ is $\color{blue}{\text{right angled}}$.   
