The Pythagorean theorem is perhaps one of the most important theorems in mathematics. This theorem allows us to relate the sides of a right triangle using an algebraic equation. There are a wide variety of proofs that can be used to prove the Pythagorean theorem. However, the most important are the proof of Pythagoras, the proof of Euclid, the proof through the use of similar triangles and the proof through the use of algebra.

Here, we will look at each of these proofs in detail.

## Proof of Pythagoras

Let’s start with the following triangle:

This triangle has legs with lengths *a* and *b* and a hypotenuse with length *c*. Now, we use four of these triangles to form a square that has sides of length $latex a+b$ as shown in the following image:

Since the hypotenuse of these triangles is equal to *c*, the sides of the inner square are also equal to *c* and their area is equal to $latex {{c}^2}$.

Now, we can also arrange the triangles as follows and form two squares that have areas $latex {{a}^2}$ and $latex {{b}^2}$.

Therefore, we know that the area of both large squares is the same in both cases. Furthermore, we also know that the four triangles are the same in both cases. This means that the area of the squares $latex {{a}^2}$ and $latex {{b}^2}$ is equal to the area of the square $latex {{c}^2}$. That is, we have:

$latex {{a}^2}+{{b}^2}={{c}^2}$

## Proof of Euclid

In the diagram below, triangle ABC is a right triangle that has a right angle at A.

According to the Pythagorean theorem, the square on side BC is equal to the sum of the squares on sides BA and AC.

We draw the line AL that goes from A and is parallel to the sides BD and CE. In addition, we also draw the lines AD and FC.

Because angle BAC and angle BAG are right angles, lines CA and AG form a straight line. For the same reason, lines BA and AH also form a straight line.

Now, since the angles DBC and FBA are right, we can add the angle ABC to each and that means that the angles DBA and FBC are equal.

Also, because segment DB equals BC, and segment FB equals BA, sides AB and BD are equal to sides FB and BC respectively. Also, the angle ABD equals the angle FBC, so the base AD equals the base FC and the triangle ABD equals the triangle FBC.

We note that the parallelogram BL is twice the triangle ABD since they share the same base BD and are on the same parallel segments BD and AL, which means that the height of the triangle is equal to the height of the parallelogram.

Also, the square GB is twice the triangle FBC since they have the same base FB and are in the same parallel segments FB and CG.

Therefore, the parallelogram BL is equal to the square GB.

Following this same process, we can form the segments AE and BK to show that the parallelogram CL is equal to the square HC. This means that the square BDEC is equal to the sum of the squares GB and HC.

Considering that the square BDEC is described by the side BC and the squares GB and HC are described by the sides BA and AC respectively, we have that the square of the side BC (hypotenuse) is equal to the sum of the squares of BA and AC ( legs).

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## Proof using similar triangles

Two triangles are similar when their corresponding angles share the same measures and their corresponding sides have the same proportions.

We use the following similar triangles:

Triangle ABD and triangle ABC have the following characteristics:

- ∠A = ∠A these angles are common
- ∠ADB = ∠ABC both angles are right angles

We see that these triangles share two angles. Furthermore, we know that all triangles have an internal sum of angles equal to 180°, which means that if two triangles have two angles with the same measures, the third angle must also have the same measures.

Therefore, we can deduce that the triangles ABD and ABC have the same angles. This means that these triangles are similar. Similarly, we can prove that the triangles BCD and ACB are also similar.

Since the triangles ABD and ACB are similar, we have the proportions $latex \frac{AD}{AB}=\frac{AB}{AC}$. We can rewrite this and say that $latex AD \times AC = {{AB}^2}$.

Similarly, the triangles BCD and ACB are similar, so we have the proportions $latex \frac{CD}{BC}= \frac{BC}{AC}$. We can rewrite this and say that $latex CD \times AC={{BC}^2}$.

Using these two equations, we can conclude that $latex {{AC}^2}= {{AB}^2}+{{BC}^2}$. We have proved the Pythagorean theorem.

## Proof using algebra

To prove the Pythagorean theorem using algebra, we have to use four copies of a right triangle that have sides *a* and *b* arranged around a central square that has sides of length *c* as shown in the diagram below.

In this diagram, *b* is the base of the triangles, *a* is the height, and *c* is the hypotenuse. By arranging the triangles as shown in the diagram, we form a large square that has sides of length $latex a+ b$.

The area of the central square formed by the hypotenuses of the triangles is equal to $latex {{c}^2}$. Also, the area of the square with sides $latex a+b$ is equal to the area of the four triangles plus the area of the central square with sides of *c*. That is, we have:

$latex {{(a+b)}^2}=4(\frac{1}{2}\times a\times b)+{{c}^2}$

$latex {{a}^2}+{{b}^2}+2ab=2ab+{{c}^2}$

$latex {{a}^2}+{{b}^2}={{c}^2}$

## See also

Interested in learning more about the Pythagorean theorem? Take a look at these pages:

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