# Proving $a^2+b^2=c^2$ where $a,b,c$ are side lengths of a right triangle.

Proving $a^2+b^2=c^2$ where $a,b,c$ are side lengths of a right triangle.

First, I have never done a proof before, sorry I am so poor here. I have spent many hours but my actions have mostly used wrong ideas. I will only say what I have done that I think is right. If you need my other writing I can write it out but it was mostly about fitting shapes inside the triangles and I think the big mistake is that if the equal statement $a^2+b^2=c^2$ were to be wrong then anything about the inside triangles cannot prove the statement to be wrong because the algebra is connected to the geometry of the triangle and the shapes are inside the geometry of the triangle.

Start: I started by drawing a square with sides $d,e,f,g$ all the same as a. The biggest possible right triangle has two sides the same length. This square can fit at least two right triangles inside it both no smaller than the other with one side equal to a. That is true because if you made one smaller than the other and pushed it into one corner of the square then the bigger triangle touching it would overlap the corners of the square which is wrong.

So then the sum $d+e+f+g=4a$ is more than $a+b+c$ where $a,b,c$ are the side lengths of one of the triangles and $4a>a+b+c$. Then I square

$$\begin{eqnarray*} 16a^2&>& (a+b+c)^2\\ &=&a^2+b^2+c^2+2(ab+bc+ac)\\ \end{eqnarray*}$$

The only thing I can do here is use the equation at the top so if $a^2+b^2=c^2$ isn’t true for the triangle then it’s $a^2+b^2=c^2+z$ for some $z$. But I subtracted that from the expression and it wasn’t wrong which I thought it would have to be since I added the $z$. I guess the extra $z$ can’t show the equation is wrong since any number could be equal to it including those like zero which it is not really equal to atleast if the equation were wrong. But if there is a rule that $z$ isn’t zero included somehow there should be a way to conclude having $z$ not be zero is wrong. But then I checked and found $a,b,c,z$ with $z$ not zero so that the inequality holds so I’m very confused. Maybe the squares are too simple and I should have chosen a rectangle or put squares around the outside of the triangle. I know a book that proves this but I want just some clues whether my actions are appropriate to solving this and otherwise a clue for a new way since I don’t care about it for a class I just like to think about the question.

Edit. I have corrected (a+b+c)^2 I think. I am also grateful for the answer below which gives a good clue and just as much as I wished for. It is certainly an interesting choice of shapes and different from what I was doing in that there are no gaps between the shapes. It helps clarify that as I am asking about ways that work I am also asking if an approach using gaps between shapes and estimations rather than exactness could work as an approach (perhaps just because I have spent some hours and wish to recover some value). Is that known? Or perhaps the notion of grouping proofs into approaches is just because the mind is simple and we can't talk about the correctness of approaches only whether one proof we are handed is right or wrong.

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A related thread. –  Guess who it is. Jan 3 '12 at 4:33
I don't think you expressed $(a + b + c)^2$ correctly. –  The Chaz 2.0 Jan 3 '12 at 5:07
How about proving via rule of cosines? –  VelvetThunder Jan 3 '12 at 5:32
Well, the poster apparently has little experience with proofs(and probably geometry) and I doubt if the OP knows about the cosine rule. :) –  Eisen Jan 3 '12 at 8:39
I would have to respectfully disagree with the rule of cosines because it would be using circular reasoning: our values from the trigonometric functions depend on the Pythagorean theorem, so stating that we should use the rule of cosines to prove the Pythagorean theorem is somewhat analogous to using the quadratic formula to show prove the method of completing the square. –  StudentofEuler2718 Aug 3 '13 at 2:59

Draw $AD\perp BC$.Now notice that $\triangle ADC \sim \triangle BAC$ where $\angle BAC=90^\circ$.From that similarity, we have $\frac{BC}{AC}=\frac{AC}{DC}$ and so $AC^2=BC\cdot DC$(Here, you need to express an angle as $\alpha$ and the other angle in that triangle excluding the right angle will now be $90^\circ-\alpha$).

I leave it to you prove that $AB^2=BC\cdot BD$.Adding up, we get $AB^2+AC^2=BC^2$.

Notice that here I am proving $a^2=b^2+c^2$ but that is because of the choice of $\angle A$ as the right angle.

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This is my favorite proof of the Pythagorean theorem. The algebra for this proof is really easy and straight forward, but it is harder to see the actual proportions that are stated in the Pythagorean theorem, if you want a proof that is clear from a visual stand point then I would look at the Lattice proof.

From a geometry standpoint it is important that you note WHY you are able to arrive to your conclusions in your proofs. In the first proof I mention, our proof is dependent on being able to show that the rhombus with sides c is in fact a square. We are able to do this because we are assuming that the sum of the angles in a triangle are 180 degrees, and that is a consequence of Euclid's 5th axiom. This is why the Pythagorean theorem fails for non-euclidean geometries.

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This is a special case of the Law of Cosines, namely one where the cosine is zero because the angle between two of the sides is 90 degrees. Go from the Law of Cosines, and when you simplify the specific case of theta = 90 degrees, a^2+b^2=c^2 will line up.

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Start with four triangles, each with sides 'a', 'b', and 'c'. Put them in a square such that side 'c' is always facing outwards. Of course, the total area of this figure is c squared. Each one of the four triangles has an area of ab / 2, therefore the four of them in total have an area of 2ab. The square in the middle, if you look closely, has sides of length b-a. Therefore, the total area of this figure is (b-a) ** 2 + 2ab. This equals a ** 2 - 2ab + b ** 2 + 2ab. The positives and negatives cancel each other out, and we are left with a squared+ b squared.

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