Question

Prove that in a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Answer 1

1) Suppose the triangle is not a right triangle.

2) Label the vertices A, B and C (There are two possibilities for the measure of angle C: less than ${90}^{\circ }$ or greater than ${90}^{\circ }$.
3) Construct a perpendicular line segment CD.
4) By the Pythagorean Theorem, $B{D}^2=a^2+b^2=c^2$, and so $BD=c$. Thus we have isosceles triangles ACD and ABD. It follows that we have congruent angles $CDA=CAD$ and $BDA=DAB$.

But this contradicts the apparent inequalities
$BDA<CDA=CAD<DAB$ or $DAB<CAD=CDA<BDA$.

Henced proved that in a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.