Proving Pythagoras by Dissection

What is to be proved

Take any right triangle. Call its two legs, the sides that meet at the right angle, a and b, and call its hypotenuse c. Pythagoras’ theorem says that $a^2+b^2=c^2$a² + b² = c².

Checking one triangle, such as $3^2+4^2=9+16=25=5^2$3² + 4² = 9 + 16 = 25 = 5², shows that the rule works for that triangle. A proof shows that it works for every right triangle, whatever the lengths a and b are. The proof below works with areas: $a^2$a² is the area of a square of side a, $b^2$b² is the area of a square of side b, and $c^2$c² is the area of a square of side c.

One right triangle, with legs a and b and hypotenuse c.

Four copies in a square

Draw a square whose sides are a + b long. Put a copy of the triangle in each of its four corners, with the right angle in the corner, so that going around the square each side is made of a leg a followed by a leg b. That uses up each side exactly, since a + b is the length of a side.

The four hypotenuses, each of length c, enclose a shape in the middle. It is tilted, and it has four sides of length c.

Four copies of the triangle in a square of side a + b. The shape left in the middle is the square on the hypotenuse, $c^2$c².

The hole is a square

Four equal sides are not enough to make a square: a square pushed over into a rhombus still has four equal sides, but its corners are no longer right angles. So the corners of the hole need checking.

First, the two acute angles of the triangle, the two corners that are not the right angle, add up to 90°. Two copies of the triangle fit together along their hypotenuses to make an a by b rectangle, and the second drawing below shows two such rectangles. At a corner of the rectangle, the hypotenuse splits the 90° corner into two angles. One is an acute angle of the first copy. The second copy is the first one turned half way round, so the angle it puts at that corner is the other acute angle of the triangle. So the two acute angles make 90° together.

Now look at a corner of the hole. It lies on a side of the big square, and a side is a straight line, which is a half turn, 180°. Three angles fill that straight line: an acute angle of the triangle on one side, the corner of the hole, and an acute angle of the triangle on the other side. The first triangle has its leg a ending there, and the second has its leg b starting there, so they bring the two different acute angles, which add to 90°. That leaves 180° − 90° = 90° for the corner of the hole.

The same is true at all four corners. The hole has four sides of length c and four right angles, so it is a square of side c, and its area is $c^2$c².

Slide the same triangles

Now move the same four triangles to new places inside the same square. Put two of them together to make an a by b rectangle in one corner, and the other two to make another a by b rectangle in the opposite corner.

What is left over is two squares. The rectangle in the top left corner is b wide and a tall. Beside it, the top edge has a + b − b = a left, and the rectangle in the bottom right corner starts a below the top edge. So the space in the top right corner is a wide and a tall: a square with area $a^2$a². In the same way, the space in the bottom left corner is b wide and b tall: a square with area $b^2$b².

The same four triangles, paired into two a by b rectangles. The two squares left over are $a^2$a² and $b^2$b².

Compare the two arrangements

Both drawings show the same big square, of side a + b, and the same four triangles. In each, the area of the big square is the four triangles plus the area left over. Take the four triangles away from the same area, and what is left must be the same both times. In the first drawing that is $c^2$c², and in the second it is $a^2+b^2$a² + b². So $a^2+b^2=c^2$a² + b² = c².

Nothing in the argument used 3 and 4, or any particular lengths. It works for a right triangle of any shape, which is what makes it a proof.

Algebra gives the same result. The big square has area ${\left(a+b\right)}^2=a^2+2ab+b^2$(a + b)² = a² + 2ab + b². Each triangle has area ½ab, so the four together have area 2ab. The first drawing says ${\left(a+b\right)}^2=2ab+c^2$(a + b)² = 2ab + c², so $a^2+2ab+b^2=2ab+c^2$a² + 2ab + b² = 2ab + c². Take 2ab from both sides: $a^2+b^2=c^2$a² + b² = c².

Where the right angle was used

The proof used the right angle twice: two copies of the triangle make a rectangle only because the corner is 90°, and the corners of the hole are right angles only because the acute angles add to 90°. Take the right angle away, and the theorem fails.

Sides of 3 and 4 with a square on every side. At 110° the square on the third side is more than 9 + 16 = 25. Turn the corner to 90°, and only there do the two smaller squares add up to the largest.

The usual mistakes

Adding the legs. The proof shows that the areas of the squares add, not the lengths of the sides. With legs 5 and 12, the hypotenuse is $\sqrt{25+144}=\sqrt{169}=13$√(25 + 144) = √169 = 13, not 5 + 12 = 17.

Using the rule on a triangle with no right angle. The proof needed the 90° corner, so the rule says nothing about other triangles: sides of 4, 5 and 6 give 16 + 25 = 41, but $6^2=36$6² = 36.