Proof 4: Liu Hui's Dissection (Cut-and-Paste)

A dissection proof attributed to Liu Hui (3rd century AD). Starting with side-by-side squares a² and b², we cut two triangles and translate them to the top to form a single tilted square c². This visually demonstrates a² + b² = c².

Prerequisites: pythagorean-theorem · Difficulty: intermediate

Liu Hui's "cut-and-paste" proof rearranges the squares and into a single square on the hypotenuse:

Strategy: place the squares (red) and (blue) side-by-side, cut two right triangles from the arrangement, and translate them upward. The pieces rejoin into a tilted square of side , and since cutting and translating preserves area, .

Step 1: Initial Squares

Start with square (blue) and square (red) placed side-by-side on a common baseline. Total area = .

Step 2: Pythagorean Theorem

Proof strategy (Liu Hui's Dissection). Cut two specific right triangles from the side-by-side arrangement and translate them upward. The pieces rejoin to form a tilted square; since cutting and translating preserves area, .

Step 3: The Cuts

Cut two specific right triangles from the arrangement:

Step 4: Transformation

Translate the triangles to the top of the figure:

Step 5: Result: Square

The pieces rejoin to form a perfect tilted square. The side length of this square is the hypotenuse . Since area is conserved: .

The pieces rejoin to form a perfect square tilted at an angle. The side length of this square is the hypotenuse . Since area is conserved when we cut and rearrange pieces:

Historical Note

This proof is attributed to Liu Hui (3rd century AD), who wrote a commentary on The Nine Chapters on the Mathematical Art. His text describes "vermilion" (red) and "indigo" (blue) squares being cut and moved according to the "In-Out Complement Principle" (出入相補) — that area is conserved when a figure is dissected and rearranged.

Similar dissection proofs were later independently developed by the Arab mathematician Thabit ibn Qurra (9th century), demonstrating the universality of this visual logic. ∎