In a right angled triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides.

I have several ways of proving this theory wrong. Nevertheless, I will just explain one of my objections.

Imagine that the triangle was ‘pasted to a sphere’s surface. Now the side view would be of a triangle of a 2D hypotenuse, instead of a 1D one. Now obviously, the curved edge would be at an angle to the straight edge of this hypotenuse. Therefore, this “ridge ’’ in the hypotenuse would be at an angle to the original 1D line, and since it is 2D, it would have a height of it’s own.

So far, Pythagoras might be giggling in Greek at me, in his grave. But now I suddenly unleash my genius at him in full blaze, blinding him, and CRUSHING HIS OWN ABILITY INTO OBLIVION WITH MY OWN….

For This height would, at least theorotically, ACT AS A COMPNENT TO THE HORIZONTAL LENGTH OF THE ORIGINAL HYPOTENUSE…

So if the original were calculated according to that old gents’ method, which does not make allowance for this extra length ion such a circumstance, theresult WOULD BE WRONG!!!

Having, as I promised, disproved This great man as though I were a smashing a snail, I will now await my next challenger…WHOM I WILL STUN AND LEAVE CRUSHED IN THE DUST..