Proof 3 of Pythagorean Theorem: Three Algebraic Proofs

Three elegant algebraic proofs using triangles arranged differently. (A) Outer square: 4 triangles around a c×c square form an (a+b)² square. (B) Inner square: 4 triangles inside a c×c square leave an (a-b)² center. (C) Gougu / Pinwheel: The complete diagram showing both arrangements simultaneously. All yield a² + b² = c² through simple algebra.

Prerequisites: pythagorean-theorem · Difficulty: intermediate

Three algebraic proofs of the Pythagorean theorem, all built from right triangles with legs $a$ , $b$ and hypotenuse $c$ (each of area $\frac{1}{2} ab$ ):

a^{2} + b^{2} = c^{2}

Strategy: arrange four triangles three ways. (A) Around a tilted $c \times c$ square to fill an $(a + b)$ square. (B) Inside a $c \times c$ square, leaving an $(a - b)$ gap. (C) Superimpose both to add the identities. Each yields $a^{2} + b^{2} = c^{2}$ by simple algebra.

Step 1: Proof A: Outer Square

Arrange 4 triangles around a tilted $c \times c$ square. The outer boundary forms an $(a + b) \times (a + b)$ square.

The algebra: $(a + b)^{2} = 4 \cdot \frac{ab}{2} + c^{2} = 2 ab + c^{2}$ .

Simplifying: $a^{2} + 2 ab + b^{2} = 2 ab + c^{2}$ , so $a^{2} + b^{2} = c^{2}$ .

Step 2: Proof B: Inner Square

Arrange 4 triangles inside the $c \times c$ square, with each hypotenuse along a side. The central hole forms an $(a - b) \times (a - b)$ square.

The algebra: $c^{2} = 4 \cdot \frac{ab}{2} + (a - b)^{2} = 2 ab + (a - b)^{2}$ .

Expanding: $c^{2} = 2 ab + a^{2} - 2 ab + b^{2} = a^{2} + b^{2}$ .

Step 3: The Combined Identities (Gougu Diagram)

Superimpose both proofs on the same diagram — the classic Gougu Diagram or "Pinwheel".

From Proof A: $c^{2} = (a + b)^{2} - 2 ab$

From Proof B: $c^{2} = (a - b)^{2} + 2 ab$

Adding the two equations: $2 c^{2} = (a + b)^{2} + (a - b)^{2}$ .

Expanding: $2 c^{2} = (a^{2} + 2 ab + b^{2}) + (a^{2} - 2 ab + b^{2}) = 2 a^{2} + 2 b^{2}$ .

Dividing by 2: $c^{2} = a^{2} + b^{2}$ .

Three arrangements, one theorem: $a^{2} + b^{2} = c^{2}$ .

The "combined identities" approach (Proof C) effectively "cancels out" the complexity of the rectangular areas ( $2 ab$ ), leaving purely the sum of squares.

Historical Context

Proof A (Outer Square) appears in many algebra textbooks and connects the binomial expansion $(a + b)^{2}$ to the Pythagorean theorem.

Proof B (Inner Square) shows the complementary relationship — working inward from $c^{2}$ instead of outward from $(a + b)^{2}$ .

Proof C (Combined / Gougu) corresponds to the Gougu Theorem derivation found in ancient Chinese mathematics (Zhoubi Suanjing) and Indian mathematics (Bhaskara II). ∎