I still maintain that this (originally from ancient China) is the clearest proof, and gives the best insight into <i>why</i> the Pythagorean Theorem holds.<p><a href="https://cdn.britannica.com/43/70143-004-CCB17706/theorem-demonstration-squares-proof-Pythagorean-b-square.jpg" rel="nofollow">https://cdn.britannica.com/43/70143-004-CCB17706/theorem-dem...</a><p>It is not immediately obvious why the area of the hypotenuse square should be equal to the sum of the areas of squares drawn on the other two sides of the triangle.<p>It is clear that the lengths of a, b and c are connected -- if we are given the length of any two of (a, b, c), and one angle, then the remaining side can only have one possible length.<p>So far, so simple; what is less clear is why the exact relationship for right triangles is c^2 = a^2 + b^2.<p>The other proofs demonstrate that the relationship holds, but give little insight.<p>The geometric proof linked above makes the relationship crystal-clear.<p>For any right triangle we can define a 'big square' with sides (a + b). <i>The hypotenuse square is simply the area of the 'big square' with 4 copies of the original triangle removed.</i><p>Simple algebra then gives us the formula for the hypotenuse square:<p>The big square has area: (a+b)^2 = a^2 + 2ab + b^2<p>The original triangle has area: ab/2<p>1 big square minus four original triangles has area: (a+b)^2 - 4ab/2 = a^2 + b^2<p>Similarly, if you take the hypotenuse square, and subtract 4 copies of the original triangle, you get a square with sides (b - a). This is trivial to prove with algebra but the geometric visualisation is quite neat, and makes clear why the hypotenuse square must always equal the sum of the other two squares.