Introduction
The Pythagorean theorem says that in a right triangle, the square of the longest side equals the sum of the squares of the other two. This reading works through one of its shortest proofs, which compares two ways of counting the area of one square. It then checks every assumption the argument made, because a proof is only as strong as the facts it rests on.
The Problem
Start with a right triangle: a triangle with one 90-degree angle. Call the two sides that form the right angle the legs, with lengths a and b. Call the longest side, opposite the right angle, the hypotenuse, with length c. The letters stand for any lengths at all, so whatever you prove about them holds for every right triangle.
The goal is to show that
a^2 + b^2 = c^2
for every right triangle. In words: the area of a square built on the hypotenuse equals the combined areas of the squares built on the two legs.
You have four identical copies of the triangle and a large square whose sides each have length a + b.
Figure: A large square with side length a + b. On each side, a mark divides the side into a piece of length a and a piece of length b, going around in the same direction. Joining neighboring marks cuts off four right triangles at the corners, each with legs a and b, and leaves a tilted square with side c in the middle. Alt text: A square with four right triangles in its corners, arranged so their long sides enclose a smaller tilted square in the center.
Working It Through
Step 1: Arrange the four triangles inside the large square
Go around the large square and mark each side so that it splits into a piece of length a followed by a piece of length b, always in the same direction. Join each mark to the mark on the next side. This cuts off one triangle at each corner of the large square.
Each corner triangle has a right angle, because the corners of a square are right angles. Its two legs have lengths a and b. So each corner triangle has the same two legs and the same angle between them as the original triangle, and it is a copy of it, with a long side of length c. The four long sides enclose a four-sided shape in the middle, tilted relative to the large square, with every side of length c.
Step 2: Find the area of the large square in two ways
The first way uses the formula for the area of a square, side times side:
\text{area} = (a + b)^2
The second way adds up the pieces. Each triangle has area one half of base times height, which for a right triangle is \tfrac12 ab. There are four of them. If the middle shape is a square of side c, its area is c^2. So the total is
\text{area} = 4 \times \tfrac12 ab + c^2 = 2ab + c^2
Both expressions describe the same area.
Step 3: Expand and simplify
Set the two expressions equal:
(a + b)^2 = 2ab + c^2
Multiplying out the left side gives (a + b)(a + b) = a^2 + 2ab + b^2. So
a^2 + 2ab + b^2 = 2ab + c^2
The term 2ab appears on both sides. Subtract it from both, and what remains is
a^2 + b^2 = c^2
A quick test with numbers: take a = 3 and b = 4, so the large square has side 7 and area 49. The four triangles have area 4 \times \tfrac12 \times 3 \times 4 = 24. The middle square must then have area 49 - 24 = 25, so c = 5. And 3^2 + 4^2 = 9 + 16 = 25, as the theorem says.
Step 4: Check which facts the argument assumed
The algebra is airtight. The geometry needs checking, because Step 2 assumed the middle shape is a square. All four of its sides have length c, but a shape with four equal sides can be a squashed diamond. It's a square only if its corners are right angles.
Look at one corner of the middle shape. It sits on a side of the large square, where two triangles meet. Along that straight side, three angles sit next to each other: an acute angle of one triangle, the corner of the middle shape, and the other acute angle of the next triangle. Angles that together make up a straight line add to 180 degrees (Euclid, Elements I.13). The two acute angles of a right triangle add to 90 degrees, because all three angles of any triangle add to 180 and the right angle takes 90 of them (Euclid, Elements I.32). So the corner of the middle shape is 180 - 90 = 90 degrees. The same holds at all four corners, so the middle shape is a square.
Step 1 also assumed a fact. It used the claim that two triangles with two equal sides and an equal angle between them are copies of each other (Euclid, Elements I.4), to conclude that each corner triangle has a long side of length c.
Step 5: Compare with Euclid's own proof
Euclid proves the same theorem as Proposition 47 of Book I, by a different route (Euclid, Elements I.47). He builds a square outward on each side of the triangle. He then draws a line from the right-angle corner, parallel to two sides of the square on the hypotenuse, which splits that square into two rectangles. Using pairs of matching triangles, he shows that one rectangle has the same area as the square on one leg, and the other rectangle the same area as the square on the other leg.
His proof uses no algebra. It cites prior propositions instead, including how to draw a parallel line (I.31) and the fact that a parallelogram is twice a triangle on the same base between the same parallels (I.41). Both proofs reach the same equation, and both rest, at some point, on facts about parallel lines.
Key Considerations
What the proof rests on. The middle shape is a square only because the two acute angles of a right triangle add to 90 degrees. That fact comes from the angle sum of a triangle. Euclid proves the angle sum using a result about angles formed when a line crosses two parallel lines (I.29), and that result is the first in the Elements to use the parallel postulate: the assumption that through a point not on a line, exactly one parallel line can be drawn. So this short proof inherits a dependence on the parallel postulate. On a curved surface such as a sphere, where that postulate fails, the theorem fails too.
The common mistake: treating the picture as the proof. A picture shows one triangle with particular proportions, and it can hide a false assumption. A drawing in which the middle shape looks square doesn't show that it is square. The proof is the reasoning in Steps 2 to 4: the two area counts, the algebra, and the argument that every corner of the middle shape is a right angle. A picture with a careful argument attached is a proof. A picture on its own is a persuasive example.
About the name. The relationship was known and used in Babylon more than a thousand years before the Greek geometers who are traditionally credited with proving it. The name records a later tradition about Pythagoras, not who first knew the fact.
Summary
Counting the area of one large square in two ways, and checking that the middle piece is a square, proves that a^2 + b^2 = c^2 for every right triangle. Here is the solution, with each line labeled by what justifies it:
- A square of side a + b holds four copies of the triangle and a middle shape with four sides of length c. (Construction; Euclid I.4)
- Each corner of the middle shape is 90 degrees, so the middle shape is a square of side c. (Angles on a straight line sum to 180, I.13; a triangle's angles sum to 180, I.32)
- (a + b)^2 = 4 \times \tfrac12 ab + c^2 (the same area counted two ways)
- a^2 + 2ab + b^2 = 2ab + c^2 (multiplying out the left side)
- a^2 + b^2 = c^2 (subtracting 2ab from both sides)
Check: with a = 3 and b = 4, the large square has area 49, the triangles 24, and the middle square 49 - 24 = 25 = 5^2 = 3^2 + 4^2.
Line 2 is where the parallel postulate enters, through the angle sum.
References
- Euclid. Elements. Edited and translated by David E. Joyce, after T. L. Heath's edition. Worcester, MA: Clark University, 1996–2013.