Proof 3 of Pythagorean Theorem: Three Algebraic Proofs

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

Prerequisites: pythagorean-theorem · Difficulty: intermediate

This proof presents three elegant algebraic approaches, all starting from a $c \times c$ square (where $c$ is the hypotenuse). Each arrangement yields $a^2 + b^2 = c^2$ through simple algebra.

The Setup

Take right triangles with short leg $a$, long leg $b$, and hypotenuse $c$. Each triangle has area $\frac{1}{2}ab$.

Proof A: The Outer Square

Arrange 4 triangles around a tilted $c \times c$ square. The triangles' legs form the outer boundary, making an outer square with side $(a + b)$.

\text{Outer area} = (a + b)^2 = 4 \times \frac{ab}{2} + c^2
    

Simplifying: $(a + b)^2 = 2ab + c^2$, so $a^2 + b^2 = c^2$.

Proof B: The Inner Square

Arrange 4 triangles inside a $c \times c$ square, with each hypotenuse along a side. The inner square (gap) has side $(a - b)$.

c^2 = 4 \times \frac{ab}{2} + (a-b)^2 = 2ab + (a-b)^2
    

Expanding: $c^2 = a^2 + b^2$.

Proof C: Combined Identities (Gougu Diagram)

Superimpose both proofs on the same diagram. From Proof A: $c^2 = (a+b)^2 - 2ab$. From Proof B: $c^2 = (a-b)^2 + 2ab$. Adding these: $2c^2 = (a+b)^2 + (a-b)^2 = 2a^2 + 2b^2$, so $c^2 = a^2 + b^2$.

The Proof

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

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

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

Step 3: The Combined Identities (Gougu Diagram)

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

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

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

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

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

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

Conclusion

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 ($2ab$), 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).