On the Pythagorean Theorem

07 Nov 2014

On the Pythagorean Theorem

In right-angled triangles, the square on the side subtending the right-angle is equal to the (sum of the) squares on the sides containing the right-angle.

Euclid’s Elements, Book 1, Proposition 47 (R. Fitzpatrick, trans.)

Figure 1: A right triangle with squares on its sides

The Pythagorean Theorem is my favorite math problem of all time. I feel so strongly about this particular bit of geometry that I have the theorem tattooed on my chest. Over my heart. In the original Greek. Yeah, I’m that kind of nerd. Most people have some vague recollection from their high school math classes that the Pythagorean Theorem is $a^2+b^2=c^2$ ; and a few even remember that the c in that equation refers to the hypotenuse of a right triangle, while the a and b refer to the other two legs. However, most of the time, people were just taught to memorize this theorem– they weren’t taught how to prove that it was actually true. Now, the Internet is full of all kinds of really clever visual proofs involving rearranging copies of the triangle in order to form the different squares, but I’m not really a huge fan of these. They make it very easy to see that the Pythagorean Theorem is true, but they don’t really make it easy to see why the Pythagorean Theorem is true. So, today, I wanted to discuss my favorite proof for the Pythagorean Theorem, which comes to us by way of Euclid’s Elements, which was the standard textbook for math in the West for around 2000 years.

In Figure 1, you’ll see that we have our triangle $\triangle ABC$ , where $\angle ABC$ is a right angle. The figures ABDE, BCFG, and ACHJ are squares.

Figure 2: Dividing the square on the hypotenuse

My first step is to draw a line from point B which is perpendicular to $\overline{HJ}$ . You’ll notice that this divides the big square, ACHJ, into two rectangles: AJKL and CHKL. Now, it’s fairly obvious that adding these two rectangles together gives you the big square, so I’m going to attempt to prove that square ABDE has the same area as rectangle AJKL, and that square BCFG has the same area as rectangle CHKL.

Figure 3: Proving ABDE = AJKL

Let’s start with square ABDE. I’m going to draw two more line segments, now, $\overline{EC}$ and $\overline{BJ}$ . You might notice that this forms two new triangles, $\triangle ACE$ and $\triangle ABJ$ . These two triangles are congruent, since $\overline{EA}=\overline{AB}$ , $\overline{AJ}=\overline{AC}$ , and $\angle EAC=\angle BAJ$ . Now, if you remember your geometry, you’ll recall that the area of a triangle is equal to its base times its height, divided by two, while the area of a rectangle is just its base times its height. Looking at $\triangle ACE$ , we can use $\overline{AE}$ as its base, and we can see easily see that its height must be equal to $\overline{AB}$ . Since square ABDE and $\triangle ACE$ share their base and height, we can conclude that square ABDE has double the area of $\triangle ACE$ . By that same token, $\triangle ABJ$ can be seen to have a base $\overline{AJ}$ and a height $\overline{AL}$ , which means that rectangle AJKL must be double the area of $\triangle ABJ$ . Since $\triangle ACE$ and $\triangle ABJ$ are equal, that means the area of square ABDE and rectangle AJKL must also be equal.

Figure 4: Proving BCFG = CHKL

Now we can repeat the exact same process with square BCFG and rectangle CHKL. First I draw $\overline{BH}$ and $\overline{AF}$ . This gives me $\triangle BCH$ and $\triangle ACF$ , which are congruent. Square BCFG shares a base and height with $\triangle ACF$ , and is therefore double $\triangle ACF$ . Rectangle CHKL shares a base and height with $\triangle BCH$ , and is therefore double $\triangle BCH$ . Since $\triangle BCH$ and $\triangle ACF$ are equal, we know that square BCFG must equal rectangle CHKL.

Figure 5: The Theorem proved

And there you have it! We proved that the squares adjacent to the right angle are equal in area to two rectangles, and we know that those rectangles add up to the area of the square opposite to the right angle. Therefore, the sum of the squares adjacent to the right angle is equal to the square opposite to the right angle.

One thought on “On the Pythagorean Theorem”

howardat58 on 2014/11/18 at 2:25 pm said:

Thanks for the “Like”. I am following you now.
Check out my previous post, on calculating Pythagorean triples.,
http://howardat58.wordpress.com/2014/11/14/pythgoras-triples-345-a-calculator/
and somewhere in the archives is an idiotically short geometric/algebraic proof of “Pythagoras” without squares. I found it:
http://howardat58.wordpress.com/2014/07/26/and-now-pythagoras-again-with-bonus/

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