[b][size=200]Proof of Pythagoras theorem c[sup]2[/sup] = a[sup]2[/sup] + b[sup]2[/sup][/size][/b]

[size=200][size=150][list][list][*]We will prove [i]the Pythagoras Theorem[/i] on the ABC triangle through a square approach.[/*][*]It is known that c [sup]2[/sup] (square ABIH) is a flat area that represents a square with a side length c on the hypotenuse of a right angle triangle, as well as a [sup]2[/sup] (square ACFG) and b [sup]2[/sup] (square BCED).[/*][*]We will prove whether the area of the square with an area of c[sup]2[/sup] (square ABIH) will be equal to the area of the squares a[sup]2[/sup] (square ACFG) and b[sup]2 [/sup](square BCED) which proves the formula [size=200][size=100][b]c[sup]2[/sup] = a[sup]2[/sup] + b[b][sup]2[/sup][/b][/b][/size][/size][/*][/list][/list][/size][/size]