Proposition 47 of 48 • Theorem

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.

The Pythagorean theorem: in any right-angled triangle, the area of the square built on the hypotenuse (the side opposite the right angle) is exactly equal to the combined areas of the squares built on the other two sides. This is the climax of Book I and one of the most famous theorems in all of mathematics.

Before You Read

Take any right triangle and build a square on each of its three sides. The claim—known for millennia, used by builders and navigators across civilizations—is that the two smaller squares together have exactly the same area as the large square on the hypotenuse. Euclid's proof of this is not the short algebraic argument you may know; it is a careful geometric argument threading through nearly every proposition in Book I.

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Basic Constructions (5)

Triangle Fundamentals (5)

Perpendiculars & Angles (5)

Exterior Angles & Inequalities (5)

Interior Triangles & Angle Copying (5)

Parallel Lines (5)

Parallel Constructions & Parallelograms (5)

Area Theorems (5)

Area Applications (4)

Grand Finale (4)

What Euclid Is Doing

Setup: Let ABC be a right-angled triangle with the right angle at A. On each of the three sides BC, CA, AB, describe a square (Proposition 46). We must prove that the square on BC (the hypotenuse) is equal to the sum of the squares on CA and AB.

Approach: Euclid's proof — sometimes called the 'windmill' proof because of the shape of the figure — works by dropping the altitude from A to the hypotenuse BC, which divides the hypotenuse square into two rectangles. He then shows that each rectangle equals the corresponding leg-square. The dividing line is constructed as AL, drawn through A parallel to the hypotenuse square's sides (Proposition 31) — geometrically it runs along the altitude from A. The method uses two pairs of congruent triangles (Proposition 4, SAS) and the relationship between parallelograms and triangles of the same base and height (Proposition 41).

Conclusion: Through A draw AL parallel to BD (Proposition 31), crossing BC and meeting DE at L — this divides the square on BC into two rectangles, BL (on base BD) and CL (on base CE). Since angles BAC and BAG are both right, CA is in a straight line with AG; similarly BA with AH (Proposition 14). Now compare triangle ABD with triangle FBC: BD = BC and FB = BA (sides of squares), and angle DBA = angle FBC, since each is a right angle plus the common angle ABC (Common Notion 2). So triangle ABD is congruent to triangle FBC (Proposition 4, SAS). Rectangle BL is double triangle ABD (same base BD, between parallels BD and AL), and square ABFG is double triangle FBC (same base FB, between parallels FB and GC) — Proposition 41. Doubles of equals are equal, so rectangle BL = square ABFG (the square on AB). By the symmetric argument (joining AE and BK), rectangle CL = square ACKH (the square on AC). Therefore the square on BC = rectangle BL + rectangle CL = square on AB + square on AC. ✓

Key Moves

Given: right triangle ABC with right angle at A.
Describe squares on all three sides: BDEC on BC, ABFG on AB, ACKH on AC (Proposition 46).
Since angles BAC and BAG are both right, CA is in a straight line with AG; similarly BA with AH (Proposition 14).
Through A draw AL parallel to BD, crossing BC and meeting DE at L — dividing the hypotenuse square into rectangles BL and CL (Proposition 31). Join AD and FC.
AB = FB, BD = BC (sides of squares), and angle DBA = angle FBC — each is angle ABC plus a right angle (Common Notion 2).
By SAS congruence (Proposition 4): triangle ABD is congruent to triangle FBC.
Rectangle BL = 2 times triangle ABD (same base BD, same parallels BD and AL; Proposition 41).
Square ABFG = 2 times triangle FBC (same base FB, same parallels FB and GC; Proposition 41).
Therefore rectangle BL = square ABFG.
By the symmetric argument (join AE and BK): rectangle CL = square ACKH.
Square BDEC = rectangle BL + rectangle CL = square ABFG + square ACKH ✓

Try It Yourself

Draw a right triangle with legs 3 and 4 units (the right angle between them) and build squares on all three sides. Cut out the two smaller squares and see whether they cover the larger one without overlap or gap. Then try a right triangle with legs of very different lengths—does the relationship still hold?

Proof Challenge

Available Justifications

Given: Right triangle ABC with right angle at A. Construct squares on each side: BDEC on BC, ABFG on AB, and ACKH on AC.

Drag justification

Draw AL through A parallel to BD (or CE), crossing BC and meeting DE at L. Draw lines AD and FC.

Drag justification

Since angle BAC and angle BAG are both right angles, CA is in a straight line with AG. Similarly, BA is in a straight line with AH.

Drag justification

Angle DBA = angle FBC, since each is a right angle of a square plus the common angle ABC.

Drag justification

DB = BC and AB = FB (sides of squares), and angle DBA = angle FBC (step above). Therefore triangle ABD is congruent to triangle FBC.

Drag justification

The rectangle on BD (between parallels BD and AL) is double triangle ABD (same base BD, between same parallels). Square ABFG is double triangle FBC (same base FB, between same parallels).

Drag justification

Since triangle ABD = triangle FBC, the rectangle on BD = square ABFG.

Drag justification

Similarly, draw lines AE and BK. By the same argument, the rectangle on CE = square ACKH.

Drag justification

Therefore square BDEC = rectangle on BD + rectangle on CE = square ABFG + square ACKH. The square on the hypotenuse equals the sum of the squares on the two legs.

Drag justification

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Curriculum Materials

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The lesson plan, student worksheet, and answer key for Proposition 47 come with the curriculum bundles.

✓Included in Advanced (Propositions 27–48)
✓or the Complete Collection (all 48)

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Why It Matters

This is the most famous theorem in mathematics and the climax of Euclid's Book I. Every proposition from 1 through 46 feeds into this result — it is the summit that the entire book has been climbing toward. The Pythagorean theorem connects the geometry of lengths to the geometry of areas, and it is the foundation of distance measurement in Euclidean space.

Going deeper

Modern connection: The Pythagorean theorem defines the Euclidean metric: d(P,Q) = sqrt((x2-x1)² + (y2-y1)²). It underpins coordinate geometry, trigonometry, linear algebra (inner product spaces), and all of physics. Every time you compute a distance, a magnitude, or a norm, the Pythagorean theorem is at work. Its generalizations include the law of cosines, the Parseval identity in Fourier analysis, and the Hilbert space structure of quantum mechanics.

Historical note: Although called the Pythagorean theorem, the result was known to the Babylonians at least a thousand years before Pythagoras (the tablet Plimpton 322, dated circa 1800 BCE, contains Pythagorean triples). What Euclid provides — and what the Greek tradition likely contributed — is the first rigorous deductive proof from axioms. Euclid's specific proof, the 'windmill' proof, is just one of hundreds of known proofs; Elisha Loomis catalogued 367 in his 1927 book. Even US President James Garfield published an original proof in 1876. The theorem's converse (Proposition 48) completes Book I, giving an if-and-only-if characterization of right angles.

Discussion Questions

Euclid's proof works by decomposing the hypotenuse square into two rectangles. There are many other proofs — by rearrangement, by similar triangles, by algebra. What makes Euclid's proof distinctive? What tools does it avoid that other proofs use?
The altitude from the right-angle vertex to the hypotenuse is the key construction. Why is this line so important? What geometric role does it play?
The Pythagorean theorem only holds in Euclidean (flat) geometry. On a sphere or in hyperbolic space, the relationship between the sides of a right triangle is different. What does this tell us about the role of the parallel postulate in the proof?
Book I has 48 propositions. Propositions 1-46 build the toolkit, and Proposition 47 is the payoff. Is this structure — long buildup to a single climactic result — a good way to organize mathematics? What are the alternatives?

Euclid's Original Proof

In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle.

Let ABC be a right-angled triangle having the angle BAC right;

I say that the square on BC is equal to the squares on BA, AC.

For let there be described on BC the square BDEC, and on BA, AC the squares GB, HC; [I.46]

through A let AL be drawn parallel to either BD or CE, [I.31]

and let AD, FC be joined.

Then, since each of the angles BAC, BAG is right, it follows that with a straight line BA, and at the point A on it, two straight lines AC, AG not lying on the same side make the adjacent angles equal to two right angles;

therefore CA is in a straight line with AG. [I.14]

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

And, since the angle DBC is equal to the angle FBA: for each is right:

let the angle ABC be added to each;

therefore the whole angle DBA is equal to the whole angle FBC. [C.N. 2]

And, since DB is equal to BC, and FB to BA, the two sides AB, BD are equal to the two sides FB, BC respectively;

and the angle ABD is equal to the angle FBC;

therefore the base AD is equal to the base FC, and the triangle ABD is equal to the triangle FBC. [I.4]

Now the parallelogram BL is double of the triangle ABD, for they have the same base BD and are in the same parallels BD, AL; [I.41]

and the square GB is double of the triangle FBC, for they again have the same base FB and are in the same parallels FB, GC. [I.41]

But the doubles of equals are equal to one another. [C.N. 1 (and C.N. 2)]

Therefore the parallelogram BL is equal to the square GB.

Similarly, if AE, BK be joined, the parallelogram CL can also be proved equal to the square HC;

therefore the whole square BDEC is equal to the two squares GB, HC. [C.N. 2]

And the square BDEC is described on BC, and the squares GB, HC on BA, AC.

Therefore the square on the side BC is equal to the squares on the sides BA, AC.

Therefore etc.

Q.E.D.

What's Next

Proposition 47 proved that a right angle produces the sum-of-squares relationship. Proposition 48 closes the circle by proving the converse: if a triangle satisfies the sum-of-squares relationship, the angle in question must be right. Together they give a complete 'if and only if' characterization of right-angled triangles—the perfect final word for Book I.