Prove that a triangle ABC is right-angled if and only if

**Solution by Steve Dinh, a.k.a. Vo Duc Dien**

Let the three side lengths of triangle ABC be a, b and c.

Apply the law of sines, we obtain

Now applying the law of cosines, we obtain

and substitute a^{2} into the above equation

(if \angle A is not right.)

Solve for cosA, we have cosA = b/c and c/b, which implies that either angle B or angle C is right.

Conversely, assume that one of the angles, say angle A, is right.

Then sin^{2}A = 1, sin^{2}B = b^{2}/a^{2} and sin^{2}C = c^{2}/a^{2}, and sin^{2}B + sin^{2}C= (b^{2}+c^{2})/a^{2} = 1.

Therefore, sin^{2}A + sin^{2}B + sin^{2}C = 2.

**Note**: another proof is available elsewhere. A slight modification of the problem was also suggested (but did not get it) by Poland for the 1967 IMO.