RE:FRAMED: Pythagorean Triple Day!

Well, not really. But 5 days ago it was! December 5th, 2013. 122 + 52 = 132

Sure, everybody’s memorized the formula using whichever formulas you like. But really, what was the logic behind this and how did Pythagoras even encounter this? It certainly works, but why? Those are the topics that I’ll be covering today.

For those unfamiliar with the Pythagorean Theorem for whatever reason or can’t seem to recall it, it states that the value of the two legs (the perpendicular lines) of a right triangle (one with a 90 degree angle)  squared and added together would equal to the side of the hypotenuse (the longest side) squared. Or, summarized into math, a² + b² = c².

Now, to be fair, the knowledge of this theorem predates even Pythagoras and there’s evidence that the Chinese, Indians, and others had reached the same conclusion without the aid of his works. He is not the only one to have known this but he’s certainly the most famous. Another interesting fact is that the Greeks did not perform math with numbers. Pythagoras had come up with this using only shapes and lines!

Pythagoras’s work was shown in Euclid’s Elements, a treatise consisting of 13 books. I will use the proof in this to show how Pythagoras drew that conclusion.

That’s better. Math is very visual. That’s why when we look at trends we tend to graph them. Ha ha. Anyway, back on topic. Here we have three perfect squares arranged in such a way that they meet at the corners and form a right triangle. The line at point A cuts the square on the hypotenuse so that the two areas which are nicely color coded for you are equal to the other two squares on the legs of the triangle. To do this proof, we must assume 4 things which are thankfully proven for you already (taken from Wikipedia.)

  1. If two triangles have two sides of the one equal to two sides of the other, each to each, and the angles included by those sides equal, then the triangles are congruent (side-angle-side.)
  2. The area of a triangle is half the area of any parallelogram on the same base and having the same altitude.
  3. The area of a rectangle is equal to the product of two adjacent sides.
  4. The area of a square is equal to the product of two of its sides (follows from 3).

I’m not going to throw the proof at you with letters and terms that are going to confuse the math out of you and so I’m going to dumb it down to a level where even I can understand it. When I feed you a series of random capital letters, just refer back to the drawing and connect the dots to see what shape I’m talking about.

These new lines that I’ve drawn have created two new triangles (points BCF and the second ABD). What’s the purpose? Well, hold on, let me get there. The perpendicular lines of the top left square and bottom left (well duh, they’re squares) are equal and so the two triangles must be equal to each other. Congruent would be a better word since equal is a bit ambiguous. The two triangles are congruent meaning that they have the exact same dimensions and area. Refer back to the second thingy that we assumed. Because of this, we can prove that rectangle BDLK has twice the area of triangle ABD. C is on the same line as A and G and so the top left triangle is twice the area of the triangle FBC. That means the pink rectangle is equal to the area of the pink square. Since it’d take a lot longer to type out and it’s already on Wikipedia for both you and me I’ll feed you the rest which is actually not that difficult to follow surprisingly.

  1. Adding these two results, AB2 + AC2 = BD × BK + KL × KC
  2. Since BD = KL, BD × BK + KL × KC = BD(BK + KC) = BD × BC
  3. Therefore AB2 + AC2 = BC2, since CBDE is a square.

Or you can be like me and use other, easier visual aids since all this writing makes it a bit difficult to follow and frankly I don’t know about you but it gives me a headache. -Tyler

Zelkof6

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