Pythagoras's Theorem

Theorem

Given any right triangle $\triangle ABC$ with $c$ as the hypotenuse, we have $a^2 + b^2 = c^2$.

Proof 1

Consider the triangle shown below.

We can extend this triangle into a square by transforming it using isometries, specifically rotations and translations.

This new figure is shown below.

This figure is clearly a square, since all the angles are right angles, and the lines connecting the corners are easily seen to be straight.

Now, let's calculate the area of this figure.

On the one hand, we can add up the area of the component parts of the square, specifically, we can add up the four triangles and the inner square.

Thus we have the area of the square to be $\displaystyle 4 \left({\frac 1 2 a b}\right) + c^2 = 2 a b + c^2$.

On the other hand, we have the area of the square as $\left({a + b}\right)^2 = a^2 + 2 a b + b^2$.

Now these two expressions have to be equal, since they both represent the area of the square.

Thus, $a^2 + 2 a b + b^2 = 2 a b + c^2 \iff a^2 + b^2 = c^2$.

$\blacksquare$

Proof 2

Let $\triangle ABC$ be a right triangle and $h_c$ the altitude from $c$.

We have

$\angle CAB \cong \angle DCB$

$\angle ABC \cong \angle ACD$

Then we have

$\triangle ADC \sim \triangle ACB \sim \triangle CDB$

Use the fact that if $\triangle XYZ \sim \triangle X'Y'Z'$ then:

$\displaystyle \frac {(XYZ)} {(X'Y'Z')} = \frac {XY^2}{X'Y'^2} = \frac {h_z^2} {h_{z'}^2} = \frac {t_z^2} {t_{z'}^2} = \ldots$

where $(XYZ)$ represents the area of $\triangle XYZ$.

This gives us:

$\displaystyle \frac {(ADC)} {(ACB)} = \frac {AC^2} {AB^2}$ and $\displaystyle \frac {(CDB)} {(ACB)} = \frac {BC^2} {AB^2}$

Taking the sum of these two equalities we obtain:

$\displaystyle \frac {(ADC)} {(ACB)} + \frac{(CDB)} {(ACB)} = \frac {BC^2} {AB^2} + \frac {AC^2} {AB^2}$

Thus:

$\displaystyle \frac{\overbrace{(ADC) + (CDB)}^{(ACB)}} {(ACB)} = \frac {BC^2 + AC^2} {AB^2}$

This gives us $\therefore AB^2 = BC^2 + AC^2$ as desired.

$\blacksquare$

Alternative Derivation

Again from the above diagram, we have:

• $\displaystyle \frac {AC}{AB} = \frac {AD}{AC}$
• $\displaystyle \frac {BC}{AB} = \frac {BD}{BC}$

using the fact that all the triangles involved are similar.

That is:

• $AC^2 = AB.AD$
• $BC^2 = AB.BD$

Adding, we now get:

• $AC^2 + BC^2 = AB.AD + AB.BD = AB (AD + BD) = AB^2$

$\blacksquare$

Proof 3

The area of the big square is $c^2$.

It is also equal to $\displaystyle 4 \frac {ab} 2 + \left({a-b}\right)^2$.

So:

	\(\displaystyle \)	\(\displaystyle c^2\)	\(=\)	\(\displaystyle 4 \frac {ab} 2 + \left({a-b}\right)^2\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 2 a b + a^2 - 2 a b + b^2\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle a^2 + b^2\)	\(\displaystyle \)

$\blacksquare$

Proof 4

Dissect the square on the left (which has area $c^2$) as shown.

Rearrange the pieces to make the two squares on the right, with areas $a^2$ and $b^2$.

As Bhāskara II Āchārya is then reported as saying:

"Behold!"

$\blacksquare$

Proof 5

The two squares both have the same area, that is, $\left({a + b}\right)^2$.

The one on the left has four triangles of area $\displaystyle \frac {ab} 2$ and a square of area $c^2$.

The one on the right has four triangles of area $\displaystyle \frac {ab} 2$ and two squares: one of area $a^2$ and one of area $b^2$.

Take away the triangles from both of the big squares and you are left with $c^2 = a^2 + b^2$.

$\blacksquare$

Algebraic Proof

We start with the algebraic definitions for sine and cosine:

• $\displaystyle \sin x = \sum_{n=0}^\infty \left({-1}\right)^n \frac {x^{2n+1}}{\left({2n+1}\right)!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$

• $\displaystyle \cos x = \sum_{n=0}^\infty \left({-1}\right)^n \frac {x^{2n}}{\left({2n}\right)!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$

From these, we derive the proof that $\cos^2 x + \sin^2 x = 1$.

Then from the Equivalence of Definitions for Sine and Cosine, we can use the geometric interpretation of sine and cosine:

• $\displaystyle \sin \theta = \frac {\text{Opposite}} {\text{Hypotenuse}}$

• $\displaystyle \cos \theta = \frac {\text{Adjacent}} {\text{Hypotenuse}}$

Let $\text{Adjacent} = a, \text{Opposite} = b, \text{Hypotenuse} = c$, as in the diagram at the top of the page.

Thus:

	\(\displaystyle \)	\(\displaystyle \cos^2 x + \sin^2 x\)	\(=\)	\(\displaystyle 1\)	\(\displaystyle \)	Sum of Squares of Sine and Cosine
	\(\displaystyle \implies\)	\(\displaystyle \left({\frac a c}\right)^2 + \left({\frac b c}\right)^2\)	\(=\)	\(\displaystyle 1\)	\(\displaystyle \)
	\(\displaystyle \implies\)	\(\displaystyle a^2 + b^2\)	\(=\)	\(\displaystyle c^2\)	\(\displaystyle \)	multiplying both sides by $c^2$

$\blacksquare$

The Classic Proof

Let $ABC$ be a right triangle whose angle $BAC$ is a right angle.

Construct squares $BDEC$ on $BC$, $ABFG$ on $AB$ and $ACKH$ on $AC$.

Construct $AL$ parallel to $BD$ (or $CE$).

Since $\angle BAC$ and $\angle BAG$ are both right angles, from Two Angles making Two Right Angles make a Straight Line it follows that $CA$ is in a straight line with $AG$.

For the same reason $BA$ is in a straight line with $AH$.

We have that $\angle DBC = \angle FBA$, because both are right angles.

We add $\angle ABC$ to each one to make $\angle FBC$ and $\angle DBA$.

By common notion 2, $\angle FBC = \angle DBA$.

By Triangle Side-Angle-Side Equality, $\triangle ABD = \triangle FBC$.

We have that the parallelogram $BDLM$ is on the same base $BD$ and between the same parallels $BD$ and $AL$ as the triangle $\triangle ABD$.

So, by Parallelogram on Same Base as Triangle has Twice its Area, the parallelogram $BDLM$ is twice the area of $\triangle ABD$.

Similarly, we have that the parallelogram $ABFG$ (which happens also to be a square) is on the same base $FB$ and between the same parallels $FB$ and $GC$ as the triangle $\triangle FBC$.

So, by Parallelogram on Same Base as Triangle has Twice its Area, the parallelogram $ABFG$ is twice the area of $\triangle FBC$.

So $BDLM = 2 \triangle ABD = 2 \triangle FBC = ABFG$.

By the same construction, we have that $CELM = 2 \triangle ACE = 2 \triangle KBC = ACKH$.

But $BDLM + CELM$ is the whole of the square $BDEC$.

Therefore the area of the square $BDEC$ is equal to the combined area of the squares $ABFG$ and $ACKH$.

$\blacksquare$

Historical Note

This is Proposition 47 of Book I of Euclid's The Elements.

This theorem is the converse of Proposition 48: Square equals Sum of Squares implies Right Triangle.

The proof he gave is the one given here as "The Classic Proof".

Source of Name

This entry was named for Pythagoras of Samos.

See also

• Pythagorean Theorem (Hilbert Space), a broad generalisation to the context of Hilbert spaces.

Sources

• George F. Simmons: Calculus Gems (1992), Chapter $\text {B}.1$

External Links

• Pythagorean Theorem and its many proofs - 92 proofs of Pythagoras's Theorem.