Two visualizing proofs of the Pythagorean theorem

This is my first ever blog post. I had some ideas on proving the Pythagorean theorem in a direct and dynamical manner, and recently I find a lot of motivation to write them down and share them on a blog(hopefully I can keep this going), so here we go. 

I have also written a paper version(the same content as this blog) which can be found at https://vixra.org/abs/2305.0047.

The Pythagorean theorem is one of the most proved theorem of all time, most of the proofs use manipulation of areas to prove that the square of the hypotenuse is indeed the sum of the squares of the two legs. I often felt disconnected with the proofs, it was one of the situations where I could prove something, but could not quite see why. So here let me present two methods of proving the Pythagorean theorem that are hopefully direct, dynamical and visualizing.  

1 The question  

The problem to solve here is quite straight forward: given two side lengths $a$ and $b$, let them form a right-angled triangle, with $a$ and $b$ the legs. Then what is the length of the hypotenuse $c$? 

2 The first method   

Although we do not know the length of $c$, we can nevertheless draw a graph containing it. Given a right-angled triangle $abc$ with c as the hypotenuse, and point $O$ as the vertex between $a$ and $c$,

let $c$ rotate clockwise about $O$, until $c$ overlaps with a(so rotate from $OD$ to $OA$). Now the trick is, during this rotation we will trace the horizontal projection of $c$, or the projection of $c$ onto the line containing $a$. Clearly, before the rotation starts, the horizontal projection of $c$ is just $a$, but as $c$ starts rotating, its horizontal projection grows larger, and when $c$ finally overlaps with $a$, its projection will be equal to itself, $c$, which is exactly what we are trying to calculate. The dotted line in the figure above shows the total horizontal increment gained from this rotation.  

2.1 The setup 

Rotation or circular movement can be approximated by some finite amount of linear motions. For example, if an object always moves in a straight line for a short distance, before readjusting its direction of motion such that the direction of motion is always perpendicular to the radius at that point, then such movement can be a good approximation for a circular motion. The more frequent it adjusts its direction of motion, the closer to a circular motion it will be. 

Now let us approximate the rotation of the hypotenuse $c$ with 5 linear motions, call it the 5-step-rotation, in the following figure, 

the same triangle $abc$ is shown with $OB$ as the hypotenuse $c$. Moreover, $BC$, $CF$, $FH$, $HK$ and $KA$ are all line segments, they represent the 5 linear motions to approximate rotation. The direction of motion is always perpendicular to the "radius'' at that point, so,

$BC \perp OB CF \perp OC FH \perp OF HK \perp OH KA \perp OK$

let us call these 5 line segments "tangent segment'' or "tangent movement''. The horizontal component and vertical component of each tangent segment are also drawn on the graph, for example, $BQ$ and $QC$ are respectively the horizontal component and vertical component of tangent segment $BC$, $CE$ and $EF$ are respectively the horizontal component and vertical component of the tangent segment $CF$, and so on. Here two very trivial facts should be noted: the horizontal component of every tangent segment is parallel to $a$, and the vertical component of every tangent segment is parallel to $b$. Now what about the lengths of these tangent segments? Well the length of these tangent segments are not equal to each other, rather they are defined such that their vertical components are equal to each other. 

$QC = EF = IH = JK = LA$

So basically every step of the rotation it always moves downward by the same amount. As the sum of these vertical components is $b$, each vertical component of these tangent segments is $b/5$.

So $OB$ took 5 linear movements to "rotate'' downward to "become'' $OA$. Of course $OA$ does not equal to $OB$ as it was not a perfect rotation, so radius is not conserved. Although, it was still a decent approximation, and the final horizontal projection, in this case would be $OA=a+BQ+CE+FI+HJ+KL$, is the approximated hypotenuse by this 5-step-rotation. 

Now let us generalize this setup by defining the number of tangent segments taken to be $N$, the vertical component of each tangent segment then will be $b/N$, and the larger is $N$ the closer it is to a circular motion. If $N$ tends to infinity, this movement of $OB$ to $OA$ becomes a perfect rotation, the infinite amount of tangent segments together become an arc, and the final horizontal projection would exactly be equal to the hypotenuse $c$.   

2.1 Find the pattern

For the general case $N$, let us go ahead and calculate the horizontal projection. To make things easier, as the downward components of all tangent segments are $b/N$, we can just write it as Δ. Now let us look at the first tangent movement,

the original horizontal projection before the first tangent movement is just $a$, then the first tangent movement adds $BQ$ to the horizontal projection. As triangle $BQC$ is similar to triangle $abc$(because $BC\perp OB$), first horizontal increment $BQ$ is, 

$BQ=\frac{b}{a}\times QC=\frac{b}{a}\times \Delta$

now after the first movement, at the new point $C$, let us denote the vertical component and horizontal component of $OC$ by $b^{\text{'}}$ and $a^{\text{'}}$ respectively,

$b^{\text{'}}=b-QC=b-\Delta$

$a^{\text{'}}=a+BQ=a+\frac{b}{a}\times \Delta$

so the horizontal projection after the first step is a'. Now let us look at the second tangent movement,

the horizontal increment $CE$ can again be obtained by similar triangles,

$CE=\frac{b^{\text{'}}}{a^{\text{'}}}\times EF=\frac{b^{\text{'}}}{a^{\text{'}}}\times \Delta$ 

again at the new point F, denote the vertical and horizontal components of $OF$ by $b^{\text{'}\text{'}}$ and $a^{\text{'}\text{'}}$, 

$b^{\text{'}\text{'}}=b^{\text{'}}-EF=b^{\text{'}}-\Delta$ 

$a^{\text{'}\text{'}}=a^{\text{'}}+CE=a^{\text{'}}+\frac{b^{\text{'}}}{a^{\text{'}}}\times \Delta$ 

so the horizontal projection after the second step is a''. Now to the third tangent movement,

by similar triangles, the third horizontal increment $FI$ is,

$FI=\frac{b^{\text{'}\text{'}}}{a^{\text{'}\text{'}}}\times IH=\frac{b^{\text{'}\text{'}}}{a^{\text{'}\text{'}}}\times \Delta$

once more at the new point H, denote the vertical and horizontal components of $OH$ by $b^{\text{'}\text{'}\text{'}}$ and $a^{\text{'}\text{'}\text{'}}$, 

$b^{\text{'}\text{'}\text{'}}=b^{\text{'}\text{'}}-IH=b^{\text{'}\text{'}}-\Delta$

$a^{\text{'}\text{'}\text{'}}=a^{\text{'}\text{'}}+FI=a^{\text{'}\text{'}}+\frac{b^{\text{'}\text{'}}}{a^{\text{'}\text{'}}}\times \Delta$

so the horizontal projection after the third tangent movement is $a^{\text{'}\text{'}\text{'}}$. No need to continue with calculating more tangent movements, one should be able to see the pattern at this stage: the next horizontal increment is always $\Delta$ times the ratio between the vertical component and horizontal component at that point. Inductively:

$(a^{\text{'}},b^{\text{'}})=(a+\frac{b}{a}\Delta ,b-\Delta )$

$(a^{\text{'}\text{'}},b^{\text{'}\text{'}})=(a^{\text{'}}+\frac{b^{\text{'}}}{a^{\text{'}}}\Delta ,b^{\text{'}}-\Delta )$

$(a^{\text{'}\text{'}\text{'}},b^{\text{'}\text{'}\text{'}})=(a^{\text{'}\text{'}}+\frac{b^{\text{'}\text{'}}}{a^{\text{'}\text{'}}}\Delta ,b^{\text{'}\text{'}}-\Delta )$

$...$

And after all $N$ steps we would get the final horizontal projection as $a^{N\text{'}}$. Now let us observe the series of these horizontal increments:

$\frac{b}{a}\Delta \frac{b^{\text{'}}}{a^{\text{'}}}\Delta \frac{b^{\text{'}\text{'}}}{a^{\text{'}\text{'}}}\Delta ...$ 

ignore the $\Delta$ for now as every term has one. The series clearly is decreasing as the numerator is decreasing while the denominator is increasing. The numerator is pretty straight forward as it is just linearly decreasing by $\Delta$ every term. While the denominator, is always the horizontal projection of the last term. In other words, in every step, a horizontal increment is added to the original horizontal projection, and this added increment makes the next horizontal increment smaller because of enlarged denominator by itself. 

2.3 The merging triangles model

Now should the behavior of the denominator in some way remind us of wrapping elastic bands on to a tube? If there are some identical elastic bands to be wrapped layer by layer onto a large tube, as more elastic bands are put onto the tube, the resulting radius of the tube becomes larger as these elastic bands add thickness to the tube. At the same time, because the radius of the tube is larger, the next elastic band must be stretched further to be put onto the tube, as a result the elastic band must be thinner, hence will increase the radius of the tube by less amount. By the same reason the next elastic bands put on will only contribute less and less radius increments.

This analogy mimic the behavior of the denominator, but we still need the numerator to decrease linearly by $\Delta$. The correct model is, the merging triangles model, that is to merge two isosceles right-angled triangles into a bigger one. Like before, we could approximate this process by an N-step approximation.

The figure above shows the N=6 case, the triangle to the left, call it the base triangle, has side length $a$, and the "triangle'' to the right has "side length'' $b$. Of course the triangle to the right is not a real triangle, it is actually made of N rectangular slices(6 slices in the figure). These slices all have width $\Delta$=b/N, and their height, from left to right, are b, b-$\Delta$, b-2$\Delta$,...,b-(N-1)$\Delta$, so the height of these slices linearly decreases by $\Delta$. To make things easier to see, let us just denote the height of these slices, from left to right by $b$, $b^{\text{'}}$, $b^{\text{'}\text{'}}$,...,$b^{(N-1)\text{'}}$, and they are connected by the relation $b^{(i+1)\text{'}}=b^{i\text{'}}-\Delta$. Now as for the problem of the "triangle'' to the right not being a real triangle but just N slices, without doing the proof in details, it should be fairly obvious that if the number of slices, or simply N, tends to infinity, all the slices become infinitesimally thin and they together form a perfect isosceles right-angled triangle with side length b, so it will be a real triangle in the limiting case.

Now to merge the two triangles together into a bigger one, we will do it slice by slice. First, merge the first slice(the tallest slice) to the base triangle. 

This slice has height $b$ but the base triangle has height $a$, to glue this slice on we will first stretch the height of the slice to be $a$, as the area of the slice must be preserved, the width of the slice must be shrunk to $(b/a)\Delta$. The resulting shape,  

is almost an isosceles right-angled triangle, apart from a little triangle missing at the top. No need to worry about it now though, just ignore it and proceed. So the base "triangle" now has side length $a+(b/a)\Delta$, and let us denote it as $a^{\text{'}}$;

Now we merge the second slice, this slice has height , and the base triangle now has side length $a+(b/a)\Delta$ or $a^{\text{'}}$, so this slice needs to be stretched from $b^{\text{'}}$ in height to $a^{\text{'}}$ in height, then glued on: 

as a consequence, the width of this slice is shrunk to $(b^{\text{'}}/a^{\text{'}})\Delta$, and the resulting "isosceles right angled triangle'' now has side length $a^{\text{'}}+(b^{\text{'}}/a^{\text{'}})\Delta$, let's call it $a^{\text{'}\text{'}}$, 

again another small triangle is missing at the top, and again we will just ignore it for now. Now add the third slice,

the third slice has height $b^{\text{'}}-\Delta$ or $b^{\text{'}\text{'}}$, and the base triangle now has side length $a^{\text{'}\text{'}}$, so the slice must be stretched from $b^{\text{'}\text{'}}$ in height to $a^{\text{'}\text{'}}$ in height, hence its width is shrinked to $(b^{\text{'}\text{'}}/a^{\text{'}\text{'}})\Delta$. After it is glued on, the resulting "triangle" now has side length $a^{\text{'}\text{'}}+(b^{\text{'}\text{'}}/a^{\text{'}\text{'}})\Delta$, let us call it $a^{\text{'}\text{'}\text{'}}$.

Needless to glue more slices in specific, the pattern should be clear by now. We continue gluing after all $N$ slices being glued on and the result would be a right-angled triangle with side length $a^{N\text{'}}$. Now review the rotating hypotenuse case, we should be able to see the variables and math expressions of this merging triangles model being exactly the same as those of the rotating hypotenuse model, in a way that the side length of the resulting right angled-triangle evolves exactly the same way as the horizontal component of the hypotenuse, which means that the horizontal projection of the hypotenuse after the N-step rotation can alternatively be found by calculating the side length of the resulting triangle, after the N-step merging. Finally, for the merging triangles model, if $N\to \infty$, then $\Delta \to 0$, and there would be no missing little triangles anymore as the area of those little triangles are of the order ${\Delta }^2$,

so in the limiting case both the triangle cut into slices and the resulting triangle become legitimate isosceles right-angled triangle, and if we denote the side length of the resulting triangle by c,

$\frac{1}{2}{a}^2+\frac{1}{2}{b}^2=\frac{1}{2}{c}^2$

or $c=\sqrt{a^2+b^2}$. On the other hand, when $N\to \infty$, the horizontal projection of the hypotenuse after the N-step-rotation, which equals to the hypotenuse itself, is also $c=\sqrt{a^2+b^2}$. What interesting about the square root operation here is that it compromises the infinite amount of merging slice actions into just one nice operation, in our case anyway. It is also probably one of the reasons why the square root of a number is usually an irrational number. Now a conclusive figure to put the two models together:

This figure shows the $N=6$ case, the triangle $abc$ is shown here, as well as the 6 tangent segments of the rotation, by a glance the 6 tangent segments together already look much like an arc. The two triangles to be merged in the merging triangle model are also drawn: along $a$, the base triangle with area $\frac{1}{2}{a}^2$ is drawn pointing downward, and the "triangle'' cut up in 6 slices is drawn to the left of $b$. For the rotation dynamics, $c$ takes 6 tangent segments to rotate clockwise to $a$, for these 6 steps, the horizontal projection of $c$ after each tangent movement are shown by the dotted lines. At the same time, the 6 slices are merged one by one onto the right side of the base triangle. One can see that in every step, the "side length" of the resulting triangle is exactly the same as the horizontal projection of the hypotenuse, which graphically shows the equivalence of these two models.

The limiting case where $N\to \infty$ it would look like:

2 The second method 

Now, if one still feels that the first method is not quite direct and intuitive enough, as one could claim we did not find the actual hypotenuse, we only found the horizontal projection of the hypotenuse after rotating it down manually, well in that case the second method might just be a better choice. The mathematical expressions for the second method will be almost the same as in the first method, the difference comes from the modelling of the dynamics. So this time we will focus more on the setup side and skip some steps that are similar to the first method.

3.1 The setup

The initial state of this method looks like this: side $a$ lies horizontally as it is, but side $b$ is 0. Then we let side $a$ be the hypotenuse and rotate upward as $b$ increases its height, and at the same time side $a$ pick up extra lengths as it rotates upward,  

and yes we will use the N step approximation again, so side $b$ will take $N$ steps, each step by $\Delta =b/N$ to grow from 0 to $b$ in height. The hypotenuse in every step is then calculated inductively(like the horizontal projection in the first method), and the hypotenuse after $N$ steps would be the desired hypotenuse for triangle $abc$, or at least an $N$-step approximation. So in the first method, we let the vertical component start from $b$ then decrease $\Delta$ per step, and become 0 at the end, while in this method it is the opposite, the vertical component starts from 0, increases $\Delta$ per step and becomes $b$ at the end. The $N=6$ case would look like: 

so $OB=a$, it is also the initial hypotenuse, and $IB=b$. Moreover, $DB=ED=...=IH=\Delta$. In each step, the hypotenuse will do a tangent movement upward, then pick up some extra length, and the next step will do exactly the same but only with an enlarged hypotenuse. To be more specific, assume at some stage of an N-step-rotation, 

the hypotenuse at that time is $OE$ or call it $a^{\text{'}\text{'}}$, and the vertical component is $EB$ or call it $b^{\text{'}\text{'}}$. Now $PE$ is the tangent segment, $PE\perp a^{\text{'}\text{'}}$ as direction of motion is always perpendicular to the radius, and PF is the extra length the hypotenuse would pick up. The new hypotenuse after this tangent movement is $OF$, which is $OP+PF$. Here we have to make an assumption that $OP=OE$(it becomes true in the limiting case), hence $OP=a^{\text{'}\text{'}}$. Now let us calculate $PF$, the increment of the hypotenuse. Different from the first method, in right-angled triangle $FPE$, $FE$ is now the hypotenuse while tangent segment $PE$ is one of the leg. By similar triangles, as $FE=\Delta$, we have $PF=\frac{b^{\text{'}\text{'}}}{a^{\text{'}\text{'}}}\Delta$, so the new hypotenuse $O\mathrm{F,}$

$OF=a^{\text{'}\text{'}}+\frac{b^{\text{'}\text{'}}}{a^{\text{'}\text{'}}}\Delta$

and the new vertical component $BF$, 

$BF=b^{\text{'}\text{'}}+\Delta$

Unsurprisingly, if we denote the new hypotenuse $OF$ as $a^{\text{'}\text{'}\text{'}}$, and the new vertical component $BF$ as $b^{\text{'}\text{'}\text{'}}$, when we do the next tangent movement the procedure would be exactly the same and would result a new hypotenuse with length $a^{\text{'}\text{'}\text{'}\text{'}}$. Lastly if we do a full $N$-step-rotation, the hypotenuse would evolve from $a$ to $a^{N\text{'}}$, and $a^{N\text{'}}$ would be the $N$-step approxination of the hypotenuse of triangle $abc$.  

Of course, one might have already noticed another problem, that is $PF$ and $OP$ are not really on the same line. $PF$ is parallel to $OE$, but $OP$ is not parallel to $OE$ as it is the hypotenuse of the right-angled triangle $OEP$. And again, in the limiting case where triangle $OEP$ is no difference from a straight line, $OE$ would be parallel to $OP$, consequently $PF$ and $OP$ would together form a line segment. 

The corresponding merging triangles model would be:

so instead of merging slices from the tallest to the lowest, it is now from the lowest to the tallest. The equivalency of these two models is again that for a given $N$, at any step, the side length of the merged triangle is exactly the same as the hypotenuse at that step. The math expressions for the merging triangles model in this method is no different from the first method, and shall be omitted. 

Finally when $N\to \infty$, there will be no bugs in either models, and the final hypotenuse, which is the hypotenuse of triangle $abc$, must be equal to the final side length of the resulting triangle, which is again $\sqrt{a^2+b^2}$.

3 Conclusions

Now the two proofs are both complete. To compare the two methods, the second method is in a sense more straightforward as we let the hypotenuse itself evolve through rotation and the end product is exactly the hypotenuse of triangle $abc$, right at where it needs to be; while in the first method we follow the horizontal projection of the hypotenuse rather than the hypotenuse itself. On the other hand the rotation in the second method is not really a classical rotation as its radius always picks up extra length in each step of the rotation, while the rotation in the first method was more standard in some ways. From the first method we find that the square root operation is in some way connected with rotation, namely connected with the change of the horizontal and vertical components swiped by an arc.

I call these two methods dynamical and direct, dynamical because things are put into motion so I am able to see how variables evolve bit by bit; direct as in the merging triangles part, I feel like we did not do anything, but rather it was just a raw and primitive way to perceive adding areas together, so it is more about a way to see things rather than a method to do things. After all the first instinct when I see $a^2+b^2$, is to think about merging two squares together. 

I hope viewers enjoy my first ever blog post and welcome to leave a comment. For further queries please contact me through sunbang2014@gmail.com. 

4 Generalizations  .

Well, the main part is officially over but welcome to stay for a little more. There are plenty of more geometry situations other than the Pythagoras that can also be modeled with our merging slices idea. Although the scope of this blog is to demonstrate the two proofs, but it never hurts to illustrate two generalizations(with the slightest twitches) from the two methods.

4.1 First generalization

To generalize the first proof a little more, let us consider a simple question: an object is continuously doing a circular motion about the origin $O(0, 0)$, it was at $C$ with coordinate $(a, b)$, some time later we measured its y-coordinate and find it was decreased by $p$, then how much did its x-coordinate change compare to $a$?

Here $CB=b$. So the object rotated from $C$ to $A$ along the circle and its y-coordinate had gone down by $p$, and we want to find the increment in horizontal projection, $DA$ or $BE$. The answer to this question is, rather than let the hypotenuse rotate all the way down, stop it at point $A$, and the increment in horizontal projection, $BE$, would be the desired result. Equivalently, in terms of merging triangles, merge from the tallest slices until the next slice is $b-p$ in height. So it is really just using part of the triangle rather than the whole triangle to merge to the base triangle,

taking the square root of the sum of the areas of the base triangle and the trapezoid, would give us the horizontal projection at $A$, then subtract it by $a$ we would get the answer(numerical answer is not the focal point here).

4.2 Second generalization

Going back to the second method, at any step of the rotation, there is always a calculated hypotenuse for some right-angled triangle. For example, at the stage that a third of the slices have been merged, the hypotenuse at that stage is $\sqrt{a^2+(b/3{)}^2}$, it actually represents the hypotenuse of a right-angled triangle with legs $a$ and $\frac{1}{3}b$. In this spirit comes the second generalization, that is to calculate a bigger hypotenuse from a smaller one,

here the starting point $a$ is not a leg of a right angled triangle anymore, but it is already the hypotenuse of a right-angled triangle. So the question is what is the length of the bigger hypotenuse $c$. To solve this, the merging process here would be to start merging from the slice with height $g$, and stop at the slice with height $g+b$, then take the square root and we will get the desired larger hypotenuse $c$. 

This generalization is in some way similar to the cosine law, the difference being we are not provided with an angle, but instead of an extra known side $g$. To discuss its relation with the cosine law would involve trigonometry functions and angles, and would be beyond the scope of this blog.