When c squared equals d squared, then a squared also b squared, and that means that a square you have there fella.

>>4878
Thicko.

This is a parallelogram, right? c and d don't look equal from the diagram.

Pythagoras, the man who used to talk to bears and claimed to have a golden thigh and be son of the god Apollo? And had a magic arrow he used to fly? Why should I believe anything he says?

>>4880
c and d can be anything, the diagram is just to show you where they are not how big they are.

>>4883

But c and d can only be equal when the parallelogram is, in fact, a square. In which case they are both the hypotenuses to a right angled triangle in which each opposite and adjacent is equal to the length of the side of a square.

>>4886
What about fucking rectangles.

>>4887
Then it is no longer a right-angle and this theorem no longer applies.

>>4888
But c and d are equal.

>>4886
Exactly the point of >>4878

>>4889
>>4890
The only rectangle this could work in is a square. Any rectangle with non-equal side lengths cannot be filled with four right-angled triangles.

>>4891
Who cares about that? I'm just saying c and d can be equal in a rectangle, not just a square.

>>4892
Okay, so? Are you being deliberately obtuse (pun intended) or do you not understand that unless it's in a square they can't be perpendicular and so won't form right-angled triangles?

Given that in >>4889 you seemed to think that c and d being equal made it still hold up I'm going for the latter, and hence you should care about my 'proof' in >>4891 because it demonstrates that this will only work in a square.

>>4893
> But c and d can only be equal when the parallelogram is, in fact, a square.
This. This is the only thing I'm critiqueing. I'm not interested in any theorums or right-angles or anything.

Just admit you fucked up OP. I'm not looking into the maths too hard but how difficult is it when you're saying c = d to post an illustration where c actually does = d?

>>4893
So now to actually get into your codswallop: When c==d, we have a rectangle. Lets say a = 3 and b = 4 and plug that into OP's formula. What do we get? c==5 and d==5.

>>4896
OP's image is predicated on the angle between C and D being 90º.

C is 2.5, not 5.

If a ≠ b and c = d then it must be a non-square rectangle. Therefore if you are not using a square, the angle between c and d cannot be 90º. In fact, none of the angles in the triangle are 90º.

If there is no 90º angle, then the Pythagorean theorem does not apply, and the formula cannot be applied.

>>4897
Pythagoras is just a special case of al-Kashi where the cosine term is zero. In the situation in the OP, the cosine terms cancel.

c = a + b
d = a - b

c^2 = a^2 + b^2 + 2a.b
d^2 = a^2 + b^2 - 2a.b

The +2a.b and -2a.b cancel each other out so it works for any angle. Isn't that pretty?

>>4897
Rectangles have right-angles.

>>4897
If a = WY and b = YZ, then c = WZ, not OZ.

WYZ is the right angle that gives you Python Gonorrhoea or whatever.

>>4900
The angles in rectangles are usually right ones, except in Savile-Epstein rectangles, where they're wrong ones.

>>4903

The sum of the squares on the diagonals is equal to the sum of the squares on the perimeter.

Why not use a ruler?

>>4905
The Greeks were a democracy.

>>4906
You've done it now, the antiquarians will be all over you for this one.

>>4907
The great thing is that next time people say "republic, not democracy" you can point out that the two are effectively translations of each other.