Pythagorean theorem: proof by rearrangement

Proof by rearrangement

Why is a² + b² = c² for every right triangle?

Four copies of the same right triangle sit inside one big square. Slide them and the empty space changes shape from c² into a² + b², without changing its area.

Middle school
High school
Right triangle
Area conservation
Proof without words
Pythagorean triples

\text{slide}

0.00

a

3.00

b

2.25

a^2 + b^2 = c^2

$c$: 3.750
$c^2$: 14.06
$a^2 + b^2$: 14.06
$(a+b)^2$: 27.56
$4\cdot\tfrac{ab}{2}$: 13.50

Guided tour

1Four triangles around c²
The big square has side $a+b$ . Four copies of the right triangle with legs $a$ and $b$ sit in its corners, and the space left in the middle is a tilted square of side $c$ : area $c^2$ .
2Slide, do not rotate
Halfway through, three triangles are on the move. Nothing is cut, stretched or turned, so the triangles cover exactly the same area as before.
3Now a² and b²
The same four triangles now pair up into two rectangles, and the space left is a square of side $a$ and a square of side $b$ . Same big square, same triangles, so the leftovers are equal: $a^2+b^2=c^2$ .
4Any right triangle
Change $a$ and $b$ and slide again; the argument never used their values. With $a=3$ , $b=4$ (scaled here by $0.6$ ) you get the triple $3^2+4^2=5^2$ .

Explanation

1
The theorem
In a right triangle with legs $a$ , $b$ and hypotenuse $c$ :
$a^2+b^2=c^2$
Read geometrically: the square built on the hypotenuse has the same area as the two squares on the legs together.
2
Why the rearrangement proves it
Both pictures are the same square of area $(a+b)^2$ holding the same four triangles of area $\frac{ab}{2}$ each. Subtracting the triangles from the square leaves
$c^2=(a+b)^2-4\cdot\tfrac{ab}{2}\qquad\text{and}\qquad a^2+b^2=(a+b)^2-4\cdot\tfrac{ab}{2}$
so the two leftovers are equal. One fact the picture relies on: the middle shape at the start really is a square, because at each of its corners the two acute angles of a triangle add to $90^\circ$ , leaving $90^\circ$ for the corner.
3
The same proof in algebra
Expand the first picture: $(a+b)^2=4\cdot\frac{ab}{2}+c^2$ , so $a^2+2ab+b^2=2ab+c^2$ . Cancel $2ab$ and the theorem drops out.
4
Where it shows up
The distance between $(x_1,y_1)$ and $(x_2,y_2)$ is $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ : the theorem on the triangle formed by the horizontal and vertical gaps. Whole-number solutions such as $3,4,5$ and $5,12,13$ are Pythagorean triples. The converse also holds: if $a^2+b^2=c^2$ , the angle opposite $c$ is a right angle.

Check yourself

Q01

A right triangle has legs $5$ and $12$ . How long is the hypotenuse?

Q02

In the rearrangement, what stays the same between the two pictures?

Q03

Is a triangle with sides $6$ , $8$ , $11$ a right triangle?

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