Proving the Pythagorean Theorem with Ptolemy's Theorem

Ptolemy's theorem says that if the four corners of a quadrilateral ABCD lie on a circle, the product of its diagonals equals the sum of the products of its opposite sides: AC · BD = AB · CD + AD · BC. Apply it to a rectangle, whose diagonals are equal, and it becomes c² = a² + b², the Pythagorean theorem.

That makes it one of the shortest proofs on the list of proofs, provided Ptolemy's theorem is already known. The catch is in that proviso, so this page also sketches how Ptolemy's theorem itself is proved and checks whether the argument secretly assumes what it sets out to prove.

Ptolemy's theorem

A cyclic quadrilateral is a four-sided figure whose four corners all lie on one circle. Label the corners A, B, C, D in order around the circle. Its sides are AB, BC, CD, DA, and its two diagonals are AC and BD. The two pairs of opposite sides are AB with CD, and AD with BC.

The theorem is named after Claudius Ptolemy (about 100–170 CE), the astronomer of Alexandria, who used it in Book I of his Almagest to build a table of chords, an early version of a sine table. The order of the corners matters: the diagonals must be the two segments that cross inside the quadrilateral.

From Ptolemy to Pythagoras: the rectangle

Every rectangle is a cyclic quadrilateral. Its two diagonals have the same length and cut each other in half, so the point where they cross is the same distance from all four corners; a circle centered there passes through every corner, and both diagonals are diameters.

Now apply Ptolemy's theorem to a rectangle ABCD with sides AB = CD = a and BC = DA = b, and diagonals AC = BD = c:

To finish the proof for an arbitrary right triangle with legs a, b and hypotenuse c, place a copy of the triangle against its hypotenuse, rotated half a turn. The two triangles form a rectangle with sides a and b, and the hypotenuse c is its diagonal. The equation above then says a² + b² = c² for the triangle. That is the whole proof, once Ptolemy's theorem is available. The rectangle diagonal calculator uses exactly this relationship.

Proof sketch of Ptolemy's theorem

The classic proof needs only two tools, both standard results of circle geometry and similarity:

• Inscribed angles: two angles whose corners are on the circle and that "look at" the same arc are equal. For example, ∠BAC = ∠BDC, because both see the arc BC.
• Similar triangles: if two angles of one triangle equal two angles of another, the triangles are similar, so their corresponding sides are in the same ratio.

1. Choose a point K on the diagonal AC so that ∠ABK = ∠DBC. In words: copy the angle between BD and BC and lay it against BA.
2. First pair of similar triangles. Triangles ABK and DBC have ∠ABK = ∠DBC by construction, and ∠BAK = ∠BDC because both are inscribed angles on the arc BC. So they are similar, and AK/DC = AB/DB, which means AK · BD = AB · CD.
3. Second pair. Since ∠ABK = ∠DBC, taking each away from ∠ABC leaves ∠KBC = ∠ABD. Also ∠BCK = ∠BDA, inscribed angles on the arc AB. So triangles KBC and ABD are similar, and KC/AD = BC/BD, which means KC · BD = AD · BC.
4. Add. Adding the two results, (AK + KC) · BD = AB · CD + AD · BC. Since K lies on AC, AK + KC = AC, and so AC · BD = AB · CD + AD · BC.

For the rectangle, the point K turns out to be the foot of the perpendicular from B to the diagonal AC, and the two similar pairs become the small triangles of the similar triangles proof. So the Ptolemy proof is, underneath, a disguised version of that classic argument, wrapped in a more general theorem.

Is the proof circular?

A proof of the Pythagorean theorem is circular if, somewhere along the way, it relies on the Pythagorean theorem itself or on something that depends on it. That is a real danger here, because many results about circles and distances do depend on it. So it is worth checking each ingredient.

• The inscribed angle theorem is proved with isosceles triangles (two radii are equal) and the fact that the angles of a triangle add to 180°. In Euclid's Elements it is Book III, Proposition 21, and its proof does not use Proposition I.47, the Pythagorean theorem.
• Similar triangles are developed from the theory of proportion and parallel lines in Book VI, again without the Pythagorean theorem.
• Constructing K only requires copying an angle, one of Euclid's basic constructions in Book I.
• The rectangle being cyclic follows from congruent triangles: a rectangle's diagonals are equal and bisect each other. No lengths are computed, so no Pythagoras is needed.

So in classical geometry the argument is not circular. Two honest caveats remain. First, some modern courses define distance with coordinates, using √(x₂ − x₁)² + (y₂ − y₁)², and then prove facts about circles and similarity from that formula. In such a setup the Pythagorean theorem is built into the definitions, and any proof of it, this one included, is really a check of consistency. Second, the argument is not independent of the similar triangles proof, as the previous section showed; it is that proof inside a larger theorem. That is why the proof is best appreciated as a beautiful special case rather than as the simplest route to a² + b² = c². For proofs that need even less machinery, see Euclid's proof or the rearrangement proof.

More gems from Ptolemy's theorem

Because it holds for every cyclic quadrilateral, Ptolemy's theorem proves far more than the Pythagorean theorem.

The golden ratio in a regular pentagon

Take four consecutive corners A, B, C, D of a regular pentagon with side s and diagonal d. They form a cyclic quadrilateral with three sides equal to s, the fourth side AD equal to d, and both diagonals equal to d.

Ptolemy's theorem gives d · d = s · s + s · d. Dividing by s² and writing x = d/s gives x² = 1 + x, whose positive solution is

the golden ratio, rounded to three decimal places. Every regular pentagon's diagonal is about 1.618 times its side.

The addition formula for sine

Apply the theorem to a quadrilateral with one diagonal along a diameter of length 1. The sides become sines and cosines of two angles α and β, and Ptolemy's equation turns into

This is how Ptolemy built his table of chords. For a proof of the Pythagorean theorem that runs through trigonometry instead, see the trigonometric proof.

Ptolemy's inequality

For a quadrilateral that is not cyclic, the equation becomes an inequality: AC · BD < AB · CD + AD · BC. Equality happens exactly when the four points lie on a circle (or, in a degenerate case, on a line). So Ptolemy's theorem also works in reverse, as a test for whether four points are concyclic, much as the converse of the Pythagorean theorem tests for a right angle.