Demystifying Pythagoras' Theorem

Finding out why a2 + b2 = c2
is as easy as twisting one square
inside another.

While I was waking up one morning, I wondered how to reconstruct the well-known Pythagorean theorem which I'd been taught at school. Maybe you've had that half-awake mood too, when an answer to a puzzle comes to you without any effort, when you can see how it's done. That was what I was trying to allow to happen in my thoughts.

I soon realised, though, that I couldn't hope to begin the original proof after more than forty years, even with the famous diagram on view in my mind's eye. Which could be how it all works out, by not remembering what you've learnt and having to start again from the beginning.

Anyway, I had the right-angled triangle and only one of the three squares, the biggest one on the triangle's longest side which is opposite the right-angle, ¬, as a mental image. By keeping that side, which is called the hypotenuse, the same length you can make one of the other sides of the triangle smaller while the third side becomes bigger, or vice versa, like this:

Then I saw that there could be four triangles around the sides of the square, all identical, three of which were missing. When I put those three back in, the mental image of another square appeared on the outside of the first one. As the two sides of the triangle became smaller or bigger while the hypotenuse stayed the same, it looked just as if the inner square was twisting round inside the outer square, while its corners touched the other one's sides. The outer square would contract or expand to fit,and every quarter-turn four identical copies of every possible right-angled triangle would be created between the two squares.

That seemed to be the only geometry needed. It then became a question of whether the algebra worked. The right-angled triangle sides are usually given the length terms a, b, and c, with c being the hypotenuse. These would appear four times each around the square, which produced a diagram like this:

Each side of the outer square connects the b side of one triangle to the a side of the next triangle. So the side length of the outer square is (a + b)2, where the superscript means "squared", which is the same as "multiplied by itself".

Next, the area of each triangle is ab/2, where ab means "a multiplied by b", and since there are four of them their combined area is 2ab.

Then the area of the inner square, which is just c2, is also equally given by the area of the outer square (a +b)2 with the area of the four triangles 2ab taken away. That is:

Area of outer square - Area of four triangles = Area of inner square

         (a + b)2                -             2ab                 =             c2

We've only got a, b, and c in the equation, so if Pythagoras was right it can't fail to give the correct answer. But where is it? First, we have to open out (a + b)2 by multiplying each length term inside the first bracket with each term inside the second bracket, and add together any terms which are the same. Recall that
(a + b)2 is two brackets multiplied together. So:

(a + b)2 = (a + b)(a + b) = a2 + ab +ba + b2 = a2 +2ab + b2

The order of multiplying together doesn't matter, so ba = ab.

We've found out what's going on. There's a 2ab built into the outer square area
(a +b)2, which can provide the 2ab area of the four triangles we need to take away, so we end up with:

Area of outer square - Area of four triangles = Area of inner square

        a2 + 2ab +b2        -         2ab  =   a2 +b2   =             c2

This proves that: a2 +b2 = c2

By ignoring the geometrical shapes of the dotted squares on the two sides of the triangle opposite its longest side (in the first diagram), I'd reduced the mental image to something a lot easier to manipulate. As if by magic, the correct answer then came out of the algebra.

The key difference between this way of solving the problem, and all the other ways the proof of Pythagoras' theorem can be rearranged, turned out to be due to starting from symmetry. I'd imagined four identical right-angled triangles positioned equally around the original square on the longest triangle side, instead of just the one triangle I'd learnt at school. It was the simplest possible proof, and if you knew basic algebra to begin with it was the fastest to work through, amounting to only three lines.

Why make something harder than you need to, when the answer is exactly the same?

© 2010 profcrumble

Comments 1 comment

mrpopo profile image

mrpopo 6 years ago from Canada

Nice solution, I always wondered how Pythagoras actually came up with this.

When I look at Wikipedia and see all of the different proofs they have for this, I'm just overwhelmed. I really do wish schools taught the simple method and then built up on them.

We all learned how to count with our fingers, after all.

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