Proof method: Let three sides be A, B and C, and the corresponding angles are Angle A, Angle B and Angle C respectively.

Let the vertical line of side C pass through point C, that is, the height of the triangle, the vertical foot is D, and the height and length are H.

Then the area of the triangle s = HC/2.

Because BD = root number (a*a-h*h)

AD= radical sign (b*b-h*h)

So AB = BD+AD = root number (a * a-h * h)+ root number (b*b-h*h).

Because ab = c

So c = radical sign (a * a-h * h)+ radical sign (b*b-h*h)

Both sides are squares;

C*c=(a*a-h*h)+(b*b-h*h)+2* radical sign [a * a * b * b-(a * a+b * b) * h * h+h * h].

Because c*c=a*a+b*b, substitute the above formula:

2* radical number [a * [A * A * B * B-C * C * H * H+H * H * H] = 2 * H * H

Both sides are squares;

a * a * b * B- c * c * h * h+h * h * h * h = h * h * h * h * h

So a * a * b * b = c * c * h * h

Prescriptions of both parties:

a*b=c*h

Because the triangle area s = c * h/2 = a * b/2.

Because a and b are two sides of a triangle,

So only right-angled triangles are possible.

That is from c * c = a * a+b * b.

Push outward into a right triangle