Why is the usage of area necessary in proving Pythagoras' Theorem?

Question

Current proofs of Pythagoras' theorem all do so through the comparison of areas of the triangles one way or another. I know of none that is able to prove the theorem using lines alone. As one of the few foundations of mathematics, the theorem holds a very important law, yet one which is too abstract for me to perceive. While the current proofs of the theorem do manage to prove it, they do not actually show why it is so, and do not explain the unique properties of lines, points and angles which give birth to this theorem. So I am wondering if there are any proofs which do not adopt the comparison of area. Otherwise, could anyone tell me why it is impossible to prove the theorem while not using area. What property of area calculation links it to the theorem and makes it the crucial key to proving it?

Answer 1

There are hundreds of proofs of the Pythagorean theorem., many involving areas, as you mention. This is because the squared terms in the theorem naturally correspond to area, particularly of squares and triangles.

There are plenty that do not involve area. Some use higher math which I'm sure you won't be interested in. The simplest I could find uses similar triangles. Check it out on Wikipedia using the link below. Look for "Proof using similar triangles."

Answer 2

Under what geometric circumstances would multiplying a times a, b times b, and c times c, or any two lengths for that matter, not be representative of area?

No matter what proof is used, the mere fact we are multiplying perpendicular lengths CAN be viewed as the area of some figure, whether it need be drawn in or not, whether you chose to view it as such, or not, whether it is necessary to view it as such or not.

Why does it bother you that the proofs rely on area? They are just as valid and logical, regardless.

Answer 3

Actually, there is a fairly simple proof that relies only on the similarity of triangles, and does not use area at all.

Proof of the Pythagorean theorem:

Let ABC be a right triangle with m∠C = 90°. Then drop a perpendicular from C to AB and label the point of intersection D. Then ∠CAD = ∠BAC and ∠ADC ≅ ∠ACB (since both ∠ADC and ∠ACB are right), so △ACB ~ △ADC by AA similarity, and thus AD/AC = AC/AB, thus AB*AD = AC². Further, we have ∠CBD = ∠ABC and ∠BDC ≅ ∠BCA (again, since both ∠BDC and ∠BCA are right), so by AA similarity, △BDC ~ △BCA, thus BD/BC = BC/BA, so BD*BA = BC².

Now, adding the inequalities, we have AB*AD + BD*BA = AC² + BC². However, AB*AD + BD*BA = AB*AD + AB*DB = AB*(AD + DB) = AB*AB = AB², so we have AB² = AC² + BC². But AB is the hypotenuse of the right triangle, and AC and BC are its legs, so this is indeed the Pythagorean equality. Q.E.D.

I've added a diagram here: http://img152.imageshack.us/img152/9535/pythagorasrk3.png

Answer 4

Well, you can't win them all. I remember that the congruent triangle proofs boiled down to if the triangles are congruent, they are congruent. The area comes in because you deal with squared lengths, which have the units of area.