After Ptolemy's theorem, Almagest, c. 150

Four Points on a Circle

Traditional · c. 150 · EuclideanDifficultyClassical

For any quadrilateral inscribed in a circle, the product of its diagonals equals the sum of the products of its two pairs of opposite sides — and a rectangle's diagonals and sides turn that rule directly into $c^{2} = a^{2} + b^{2}$ .

Draw a circle, and inscribe a rectangle in it — a quadrilateral $A B C D$ with all four corners on the circle, one side $a$ and the next $b$ .
A rectangle's two diagonals are equal — each runs from one corner straight through the center to the opposite one. Call that shared length $c$ , without assuming anything yet about what it equals.
The sides come in two matching pairs: $A B$ and $C D$ both measure $a$ ; $B C$ and $D A$ both measure $b$ .
Ptolemy's theorem holds for any quadrilateral inscribed in a circle, rectangle or not: the product of the two diagonals equals the sum of the products of the two pairs of opposite sides.
Put this rectangle's own lengths in: $A C \cdot B D = A B \cdot C D + B C \cdot D A$ becomes $c \cdot c = a \cdot a + b \cdot b$ .
That's $c^{2} = a^{2} + b^{2}$ — reached without ever assuming it. Ptolemy's theorem has its own proof, by constructing one auxiliary point and comparing similar triangles, and that argument never mentions a right angle or the Pythagorean theorem at all.

Therefore c² = a² + b². ∎

It’s worth being precise about why this isn’t circular. Ptolemy’s theorem is proved independently — the standard argument marks a point on one diagonal so that two angles match, builds a pair of similar triangles from it, and reads off the relation among the four sides and two diagonals. Nowhere in that construction does a right angle, or the Pythagorean theorem, ever appear. Only afterward do we choose to feed it a rectangle.