Eric's Pythagorean Theorem Proof

Fig. 1 and Fig. 2 have the same area a²+b², and they are square, so they have the same edges in each and together. a²+0²=c² and 0²+0²=0² are trivial.

Fig. 3 and 4 suggest c=a+x and c=b+y.

c will fit in the square because a and b fit in the square (fig. 5). WLOG (without loss of generality) c can connect edges. We can connect adjacent edges like fig. 6, have an edge like fig. 7, or connect opposite edges like fig. 8. In the case of figure 6 we can have those edges be a and b.

In fig. 6 we want each triangle to be a and b and to be the same size so it stretches around perfectly. Note that c² < the total square area. This means that c² < a² + b² but we need to prove that impossible. See fig. 9. BC will be y. GH will be x so that CD will be x. So BC= a and y and CD = b and x. Note that x² < b² and y² < a² so that x²+y² < a²+b² but they are also equal. a²+b² < a² + b², a contradiction. Then c² is not less than a²+b²

Next we try fig. 10. This time it stretches past the length of an edge of the big square. One side will be a+x and b+y, so it will be bigger than a and bigger than b. That means we don’t have to consider it since it cannot be a and b. Then c² is not greater than a²+b²

The only thing left is that c is a side of the big square or parallel to it.

c² = a²+b² or if we look at (a+x)² and (b+y)² together we also get 2c² = 2a²+2b². It will look like figs. 11 and 12. Note that in fig. 12 it won’t be the same as like fig. 9 (it won’t stretch around evenly).

(b^2=2ax+x^2 and a^2 = 2by+y^2)

also at www.allmynoodles.com/pythagorean-theorem

I think it fails because in fig. 9 I confuse a+b with a+x.

Or maybe that's correct I don't know.

Actually I think it works!

Well I hope someone can tell me if they think it works.

Insertion:

Next we try fig. 10. This time it stretches past the length of an edge of the big square. One side will be a+x and b+y, so it will be bigger than a and bigger than b. That means we don’t have to consider it since it cannot be a and b.

What if we bend that c to fit on adjacent sides?

Using fig. 9,we have the same problem. So c² is not greater than a²+b². We can also try to bend a c that fits two adjacent sides to across and then the problem with going across occurs.

Then c² is not greater than a²+b²

The only thing left is that c is a side of the big square or parallel to it.

I have a better explained version. I would like feedback:

The arguments are generally next to each other.

In fig. 1, x is defined so that ax+ax+x² = b², and in fig. 2 y is defined so that by+by+y²=a².

Fig. 1 and Fig. 2 have the same area a²+b², and they are square, so they have the same edges in each and together. a²+0²=c² and 0²+0²=0² are trivial. The edges are all a+x and b+y.

Fig. 3 and 4 suggest c=a+x and c=b+y, but you can ignore this.

c will fit in the square because a and b fit in the square (fig. 5). WLOG (without loss of generality) c can connect edges. We can connect adjacent edges like fig. 6, have an edge like fig. 7, connect opposite edges like fig. 8, or have part of an edge like fig. 13. In the case of figure 6 we can have those edges be a and b. Usually the edges c connects make a and b to the point of c (like triangle ABH in figure 9).

In fig. 6 we want each triangle to have lengths a and b and to be the same size so it stretches around perfectly. See fig. 9. BH will have slope -a/b and BD b/a so they are perpendicular. They are also congruent. BC, DE and FG are a and CD, EF and GH are b. This will create c² as the middle square and c²< the total square area (because there’s space left over), so we will be able to check if c² < a² + b² is possible. BC will be y because AC is b+y and AB is b. Likewise GH will be x so that CD will be x. So BC= a and y and CD = b and x. Note that x² < b² (because b² = ax+ax+x² ) and y² < a² (because a² = by+by+y²) so that x²+y² < a²+b² but they are also equal (because BC = a and y and CD = b and x like we mentioned two lines up). a²+b² < a² + b², a contradiction. Then c² is not less than a²+b².

Next we try fig. 10. This time it stretches past the length of an edge of the big square. One side will be a+x and b+y, so it will be bigger than a and bigger than b. That means we don’t have to consider it since it cannot be a and b.

Then c² is not greater than a²+b².

We don’t normally have to worry about bending c so that it’s in a different category or different; all cs will be tried.

The only thing left is that c is a side or partial side of the big square or parallel to it.

In fig. 13 c is a part of a side of the square. If c < a+x it could be less than a or b. Or importantly we can show that it won’t fit going across (adjacent sides) with the same a and b as normal legs. c might not match and a and b but it has to but it doesn’t but we don’t know that, so we apply the same argument with fig. 9. c² will be given as < a²+b² because that is the whole square and c has a length less than it’s side (c < a+x and a²+b² we should know is (a+x)² see fig. 1 and fig. 2).

All that’s left is c = a+x = b+y, one complete side. c² = a²+b² or if we look at (a+x)² and (b+y)² together we also get 2c² = 2a²+2b² (c² = ax+ax+x²+a² = a²+b² and c² = by+by+y²+b² = a²+b²). It will look like figs. 11 and 12. Note that in fig. 12 it won’t be the same as like fig. 9 (it won’t stretch around evenly).

I would like feedback.

Using slopes to prove things are perpendicular isn't a great idea in geometry. I would use that the three angles of a Euclidean triangle add up to a straight line - a fact equivalent to the 5th postulate - instead. The definition of being perpendicular is forming a right angle. Negative reciprocal slopes is just a consequence of that, and since you didn't define coordinates, you technically don't have any slopes either.

Figures 3 and 4 don't suggest c = a+x or c = b+y. These would be isosceles triangles, not right triangles, and not congruent to each other. You haven't demonstrated why they should be.

Early on, it seems your square has side length a+x or b+y, but then you change it to side length a+b. Starting with a square of side length a+b gives rise to a much simpler proof, a very common one, showing (a+b)^2 = c^2 + 4(ab/2), simplifying to a^2+b^2=c^2, without the need for adding x and y to complicate the situation.

Either way, your steps aren't clear and it's hard to follow exactly what you're doing. But I'm glad you're trying this and seeking to learn.

It doesn't matter; they're at right angles.

And you can ignore fig. 3 and 4.

And once we get a+b, then we fit it to a+x.

Be careful when you judge the a+b to fit with a+x and b+y. It's tricky at least.

Basically I'm saying is if c^2 < area of square and area of square is a^2+b^2 there's a contradiction.

c will be less than the square. Without changing a and b, we make the square fit to a^2+b^2, It leads to a contradiction co c^2 can't be lesss thatn a^2+b^2l

A revised version:

The arguments are generally next to each other.

In fig. 1, x is defined so that ax+ax+x² = b², and in fig. 2 y is defined so that by+by+y²=a².

Fig. 1 and Fig. 2 have the same area a²+b², and they are square, so they have the same edges in each and together. a²+0²=c² and 0²+0²=0² are trivial. The edges are all a+x and b+y.

Fig. 3 and 4 suggest c=a+x and c=b+y, but you can ignore this.

c will fit in the square because a and b fit in the square (fig. 5). WLOG (without loss of generality) c can connect edges. We can connect adjacent edges like fig. 6, have an edge like fig. 7, connect opposite edges like fig. 8, or have part of an edge like fig. 13. In the case of figure 6 we can have those edges be a and b. Usually the edges c connects make a and b to the point of c (like triangle ABH in figure 9).

For fig. 6, we can try 2 techniques in the next 2 paragraphs. I am confident the second one works, but I don’t want to erase the first.

In fig. 6 we want each triangle to have lengths a and b and to be the same size so it stretches around perfectly. See fig. 9. BH will have slope -a/b and BD b/a so they are perpendicular. They are also congruent. BC, DE and FG are a and CD, EF and GH are b. This will create c² as the middle square and c²< the total square area (because there’s space left over), so we will be able to check if c² < a² + b² is possible. BC will be y because AC is b+y and AB is b. Likewise GH will be x so that CD will be x. So BC= a and y and CD = b and x. Note that x² < b² (because b² = ax+ax+x² ) and y² < a² (because a² = by+by+y²) so that x²+y² < a²+b² but they are also equal (because BC = a and y and CD = b and x like we mentioned two lines up). a²+b² < a² + b², a contradiction. Then c² is not less than a²+b².

The second technique also involves fig. 6. c will form a square in the square, and we can allow the larger square to be a²+b² by having the sides be a+x and b+y. Then we can test c²<a²+b². Because the c square is symmetric, all sides are a and b together as well. Hence b=x and a=y. We have the same contradiction since x² < b² (because b² = ax+ax+x² ) and y² < a² (because a² = by+by+y²) so that x²+y² < a²+b² but they are also equal. This shows that we cannot have c²<a²+b².

Next we try fig. 10. This time it stretches past the length of an edge of the big square. One side will be a+x and b+y, so it will be bigger than a and bigger than b. That means we don’t have to consider it since it cannot be a and b.

Then c² is not greater than a²+b².

We don’t normally have to worry about bending c so that it’s in a different category or different; all cs will be tried.

The only thing left is that c is a side or partial side of the big square or parallel to it.

In fig. 13 c is a part of a side of the square. If c < a+x it could be less than a or b. Or importantly we can show that it won’t fit going across (adjacent sides) with the same a and b as normal legs. c might not match and a and b but it has to but it doesn’t but we don’t know that, so we apply the same argument with fig. 9. c² will be given as < a²+b² because that is the whole square and c has a length less than it’s side (c < a+x and a²+b² we should know is (a+x)² see fig. 1 and fig. 2).

All that’s left is c = a+x = b+y, one complete side. c² = a²+b² or if we look at (a+x)² and (b+y)² together we also get 2c² = 2a²+2b² (c² = ax+ax+x²+a² = a²+b² and c² = by+by+y²+b² = a²+b²). It will look like figs. 11 and 12. Note that in fig. 12 it won’t be the same as like fig. 9 (it won’t stretch around evenly).