The Pythagorean Theorem with Semicircles and Other Shapes

Yes — the Pythagorean theorem works with semicircles, equilateral triangles and any other shape, as long as the three shapes are similar and each is built on its side in the same way. Then the areas on the two legs add up to the area on the hypotenuse. Euclid proved this version as Proposition 31 of Book VI of the Elements.

The usual statement, a² + b² = c², is about the areas of three squares (see what the Pythagorean theorem says). This page shows that the squares can be swapped for any shape you like, explains the single idea that makes it true, and works through the classic semicircle and triangle problems that appear in textbooks and contests.

Euclid VI.31: the theorem for any similar figures

In Book I of the Elements, Euclid proves the theorem for squares (Proposition I.47, explained on our Euclid's proof page). In Book VI, after developing the theory of similar figures, he returns to it with a far more general statement. In David Joyce's translation:

In right-angled triangles the figure on the side opposite the right angle equals the sum of the similar and similarly described figures on the sides containing the right angle. — Euclid, Elements VI.31

Two words do all the work. Similar means the figures have the same shape and differ only in size: one is a scaled copy of another. Similarly described means each figure is attached to its side by the corresponding edge — if a semicircle sits on the leg with its diameter along the leg, the other two semicircles must also sit with their diameters along their sides.

Strictly, Euclid's Book VI deals with polygons ("rectilinear figures"). Curved shapes need one more fact, Proposition XII.2: circles are to one another as the squares on their diameters, which Euclid proves with the method of exhaustion. That is enough to cover semicircles, quarter circles and any shape built from circular arcs.

Why it works: area scales with the square of length

Enlarge a shape by a scale factor r — every length multiplied by r — and its area is multiplied by r². Double the sides of a triangle and four copies of the original fit inside it; triple them and nine copies fit.

Now fix a shape and build it on a side of length s. Because every copy is a scaled version of one model, its area is always some fixed number times s²:

The constant k depends only on the shape and how it is attached, never on the size. For a square k = 1. For a semicircle whose diameter is the side, the radius is s/2, so the area is 1 over 2π(s/2)² = π over 8s² and k = π over 8.

Put the same shape on all three sides of a right triangle. The ordinary theorem says a² + b² = c²; multiply both sides by k:

That one line is the whole proof. The area on the hypotenuse equals the sum of the areas on the legs, whatever the shape. It also shows why the figures must be similar: the argument needs the same k on all three sides.

Area constants for common shapes

Any of these, used on all three sides, gives a true "Pythagorean theorem". So does a shape with no name at all — an outline of your hand, scaled so that the same two fingertips touch the ends of each side.

A proof that uses the triangle itself

The general version is not just a corollary; it can also prove the ordinary theorem. The trick is to choose, as the similar figure on each side, a copy of the right triangle itself, with that side as its hypotenuse.

Drop the altitude from the right angle C to the hypotenuse. It splits triangle ABC into two smaller triangles, and both are similar to ABC (each shares an acute angle with it and has a right angle). Their hypotenuses are a and b, and the big triangle's hypotenuse is c. So their areas are ka², kb² and kc² for one constant k. The two pieces fill the whole triangle exactly, so

This is essentially Euclid's own route to VI.31, and it is the argument often credited to the young Albert Einstein — see the Einstein proof and the similar triangles proof. The same altitude produces the geometric mean theorem.

Worked examples with semicircles and triangles

The lunes of Hippocrates

The most famous application of the semicircle version is about 2,400 years old. Hippocrates of Chios (active around 440 BCE, not to be confused with the physician Hippocrates of Kos) found areas of lunes — crescent shapes bounded by two circular arcs. His book is lost; we know the work through Eudemus's History of Geometry, as quoted by Simplicius in the 6th century CE.

Here is the version for any right triangle. Draw the semicircle on the hypotenuse on the same side as the triangle; by Thales' theorem it passes through the right-angle vertex. Then draw semicircles outward on the two legs. The two crescents that stick out beyond the big arc together have exactly the same area as the triangle.

The proof is pure bookkeeping with VI.31. The two leg semicircles plus the triangle cover the same region as the hypotenuse semicircle plus the two lunes:

Since π over 8a² + π over 8b² = π over 8c², those terms cancel and the lunes equal the triangle. For the 3-4-5 triangle in the figure the two lunes measure about 2.51 and 3.49 square units, and the triangle measures 1 over 2 · 3 · 4 = 6.

Hippocrates himself worked with the special case of an isosceles right triangle (the 45-45-90 triangle), where each lune equals half the triangle. The general right-triangle version is often called the lunes of Alhazen, after Ibn al-Haytham (c. 965–c. 1040), who studied it. The result mattered because Hippocrates had found a curved region whose area equals that of a polygon, which raised hopes of "squaring the circle" — building a square with the same area as a circle using only straightedge and compass. That turned out to be impossible, as Ferdinand von Lindemann proved in 1882 by showing that π is transcendental. See the history of the Pythagorean theorem for the wider story.

When the general version fails

• Shapes that are not similar. A square on one leg, a semicircle on the other and an equilateral triangle on the hypotenuse have different constants k, so nothing adds up.
• Similar shapes attached differently. A 1-by-2 rectangle attached by its long edge to the legs but by its short edge to the hypotenuse is not "similarly described". Its area is 1 over 2s² in one position and 2s² in the other.
• Triangles that are not right. The general version is exactly as strong as the original: for an acute or obtuse triangle the areas on the two shorter sides add up to more or less than the area on the longest. Conversely, if similar figures on the three sides satisfy "two equal the third", then a² + b² = c² and the triangle is right, by the converse of the Pythagorean theorem.
• Volumes. Solids scale with the cube of length, so cubes on the sides do not work: 3³ + 4³ = 91, not 5³ = 125. In fact a³ + b³ < c³ for every right triangle, because a³ + b³ = a · a² + b · b² < c · a² + c · b² = c³. For whole numbers the equation a³ + b³ = c³ has no solutions at all, the first case of Fermat's last theorem.

Areas do have a genuine three-dimensional version — de Gua's theorem for a corner of a box — and lengths generalize to any number of dimensions, as shown on the Pythagorean theorem in n dimensions. Repeating squares on the sides of ever-smaller right triangles produces the Pythagoras tree fractal.