The figure rotates onto itself

2026-05-11

Take the right triangle and its three squares. Notice the gap between the two leg-squares - it's itself a triangle, congruent to the original. Now rotate the whole upper half by 180° around the centre of the c²-square. It lands on a mirror image of itself, exactly.

That symmetry is the whole proof. The figure is its own mirror image, rotated around the centre of the hypotenuse-square. The upper half - the original triangle, both leg-squares, and the gap triangle - is congruent to the lower half. So they have the same area.

Counting twice

The whole figure can be measured two ways.

Count it as (upper half) + (c²-square) + (lower half). The upper half = a² + b² + 2 triangles (the original plus the gap, both with area ab/2). The lower half is its mirror, so it has the same composition. Total:

(a² + b² + ab) + c² + (a² + b² + ab)

Or count it as a single hexagon-and-its-mirror with 4 triangles tucked along the diagonals: the four triangles together total 2ab, and the rest is a single 2c² × something - actually we can spare ourselves the bookkeeping. Just observe: the upper and lower halves are congruent, and the upper half contains a² + b² in squares while the lower half, by symmetry, contains c²-worth of square along with two triangles whose area is ab. Equate the two:

a² + b² + ab  =  c² + ab

The ab on both sides is the same pair of triangles, just counted in different positions. They cancel. What's left is the theorem.

Why this proof feels different

Most proofs of Pythagoras compute an area two ways and equate. This one constructs a figure that cannot have any answer other than the theorem - the symmetry forces it. The proof works by making the equality visible as rotation: do the rotation, and the picture is its own answer.

Leonardo's interest in geometry was almost ornamental - he loved figures with rotational symmetry. He found this proof while illustrating Luca Pacioli's De Divina Proportione. The argument is older than him; the elegant figure is his.

Same figure. Same area. Different counts. The cross-terms cancel, and the theorem walks out.

Other proofs of a² + b² = c²: Bhāskara's rearrangement, Garfield's trapezoid, Euclid's windmill, similar triangles, Perigal's dissection.