Euclid's Proof of the Pythagorean Theorem

Euclid's proof of the Pythagorean theorem is Proposition 47 in Book I of the Elements, written about 300 BCE. A line from the right angle splits the square on the hypotenuse into two rectangles. Pairs of congruent triangles then show that each rectangle equals the square on one leg, so the two leg squares together equal the hypotenuse square.

The finished figure looks a little like a windmill, so this is often called the windmill proof. Every step uses an idea from a first geometry course: congruent triangles, parallel lines and the area of a triangle.

The statement and Euclid's figure

In D. E. Joyce's translation, Euclid states the proposition like this:

In right-angled triangles the square on the side opposite the right angle equals the sum of the squares on the sides containing the right angle.

This is a statement about areas of squares. Euclid never writes a² + b² = c²; for him "the square on BC" is a real square drawn on BC, and the formula is our modern shorthand.

Euclid puts the right angle at A, so the hypotenuse is BC. We keep his letters so you can read his text alongside this page. In the site's usual colors, the legs AB and AC are blue and orange and the hypotenuse BC is green.

The construction takes three moves:

1. Draw the three squares outward from the triangle: BDEC on BC, ABFG on AB and ACKH on AC. (Proposition I.46 shows how.)
2. Draw AL through A parallel to BD, meeting BC at M and DE at L (Proposition I.31). Because BD is perpendicular to BC, so is AL: it is the altitude of the triangle, extended through the big square.
3. Join AD and FC. Later, for the second half of the proof, join AE and BK as well.

The plan: split the big square in two

The line AL cuts the hypotenuse square into two rectangles, BMLD next to B and MCEL next to C. Euclid's whole proof comes down to two claims:

Add the two claims and the theorem follows, because the rectangles together make up the square on BC.

Test the claims on the 3-4-5 triangle first. With AB = 3, AC = 4 and BC = 5, the point M lands 1.8 from B (it is always AB²/BC; the similar triangles proof shows why). Rectangle BMLD is 1.8 by 5, area 9, which is 3²; rectangle MCEL is 3.2 by 5, area 16, which is 4². The proof shows this happens for every right triangle.

The proof in three steps

Step 1: two congruent triangles

Compare triangle FBC (the thin line FC closes it) with triangle ABD (closed by AD). They are congruent by side-angle-side, Euclid's Proposition I.4:

• FB = AB, because both are sides of the square ABFG.
• BC = BD, because both are sides of the square BDEC.
• ∠FBC = ∠ABD. Each one is a right angle plus the angle ABC: ∠FBC = ∠FBA + ∠ABC and ∠ABD = ∠CBD + ∠ABC.

One more fact is needed before step 2. Angle BAC is a right angle, and angle BAG is a corner of the square, so the two together make two right angles. By Proposition I.14 that means G, A and C lie on one straight line. In the same way, B, A and H lie on one straight line.

Step 2: each triangle is half of a four-sided figure

Euclid's Proposition I.41 says that if a parallelogram and a triangle have the same base and lie between the same two parallel lines, the parallelogram is double the triangle. In modern terms, both have the same height h over the same base s, so the parallelogram has area sh and the triangle has area 1 over 2sh. Squares and rectangles are parallelograms, so the rule applies to both of our pairs:

• Triangle FBC and square ABFG share the base FB. The square's opposite side GA lies on the line through G, A and C, which is parallel to FB, and the triangle's third corner C sits on that same line. So the square is double the triangle.
• Triangle ABD and rectangle BMLD share the base BD. The rectangle's opposite side ML lies on the line AL, which is parallel to BD, and the triangle's third corner A sits on that line. So the rectangle is double the triangle.

Step 3: put the pieces together

Chain the facts from steps 1 and 2:

The middle equality holds because congruent triangles have equal areas. The other side works the same way, using AE and BK. Triangles ACE and KCB are congruent (CA = CK, CE = CB, and each angle at C is a right angle plus angle ACB). Triangle KCB is half the square ACKH, since B lies on the line through H and A. Triangle ACE is half the rectangle MCEL, since A lies on the line AL. So square ACKH equals rectangle MCEL, and

Watch the areas move: the shear animation

Euclid's argument is static, but you can watch it as motion. A shear slides one side of a parallelogram along its own line while the opposite side stays put; base and height stay the same, so the area does too. Three moves carry each leg square onto its rectangle:

1. Shear the square ABFG along the line GAC, keeping the side FB fixed, until A reaches C. The result is a parallelogram with sides FB and BC, exactly double triangle FBC.
2. Turn it a quarter turn about B. F goes to A and C goes to D, just as in step 1. The parallelogram now has sides BA and BD and is double triangle ABD.
3. Shear again along the line AL, keeping BD fixed, until A reaches M. The parallelogram becomes the rectangle BMLD.

The animation contains the proof: the first parallelogram is double triangle FBC, the turned one is double triangle ABD, and the last shape is the rectangle.

The complete proof, written out

Here is the whole argument in the two-column format used in many geometry classes. Proposition numbers refer to Book I of the Elements.

Euclid's converse: Proposition I.48

Book I ends with the converse. If the square on one side of a triangle equals the squares on the other two sides together, then the angle between those two sides is a right angle. Euclid proves it using I.47 itself:

1. Suppose BC² = BA² + AC². At A, draw AD perpendicular to AC with AD = AB, and join DC.
2. Triangle DAC has a right angle at A, so by I.47, DC² = DA² + AC² = BA² + AC² = BC². Therefore DC = BC.
3. Triangles DAC and BAC have DA = BA, the side AC in common, and DC = BC, so they are congruent (side-side-side, I.8). The angle BAC therefore equals the right angle DAC.

The converse is what lets you test whether a triangle is right-angled just from its side lengths. See the converse of the Pythagorean theorem for worked examples, or try the right triangle checker.

Why Euclid proved it this way

The theorem was known long before Euclid. Babylonian scribes worked with Pythagorean triples more than a thousand years earlier, on tablets such as Plimpton 322, and later Greek writers credited the theorem to Pythagoras. What the Elements contributed was this particular proof. The commentator Proclus, writing in the 5th century CE, credits it to Euclid himself, and says he admired the theorem's discoverers but marveled even more at the author of the Elements for the clarity of the proof.

Historians think earlier proofs most likely used proportion and similar triangles, which Euclid develops only in Books V and VI. The windmill proof needs nothing but congruent triangles and parallel lines, so the theorem can close Book I, ready for later books to use. Euclid returns to it twice:

• Book II, Propositions 12 and 13, handle obtuse and acute triangles. They are geometric forms of the law of cosines.
• Book VI, Proposition 31, proves the theorem for any similar figures drawn on the three sides, not just squares. That is the idea behind the theorem for other shapes and behind Einstein's proof.

For the wider story, see the history of the Pythagorean theorem, or compare this proof with the much shorter rearrangement proof on the proofs overview.