Proof 2 of Pythagorean Theorem: The Puzzle Shift

A visual dissection proof showing two ways to fill an (a+b)×(a+b) square. Both use four identical right triangles (legs a and b). The remaining space is a² + b² in one arrangement, c² in the other. Therefore a² + b² = c².

Prerequisites: pythagorean-theorem · Difficulty: beginner

We want to prove the Pythagorean theorem:

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

Strategy: treat it as a puzzle. Fill an $(a + b) \times (a + b)$ square in two ways, both using the same four right triangles (legs $a$ , $b$ , hypotenuse $c$ ). One arrangement leaves a green $a \times a$ square and a red $b \times b$ square; the other leaves a single tilted blue $c \times c$ square. Since the triangles are identical, the leftover spaces must be equal.

Step 1: The $(a + b) \times (a + b)$ Square

Consider a square with side length $(a + b)$ . We'll show two different ways to fill this square, both using four identical right triangles.

Step 2: Configuration 1: Two Squares + Four Triangles

Fill the big square with a green $a \times a$ square, a red $b \times b$ square, and four right triangles. The two smaller squares sit in opposite corners, while the four triangles fill the remaining L-shaped regions.

Total area = $a^{2} + b^{2} +$ (area of four triangles).

Step 3: Pythagorean Theorem

Proof strategy (puzzle shift). Conservation of area: the same four triangles can be rearranged inside the $(a + b) \times (a + b)$ outer square to leave either the $a^{2} + b^{2}$ region (Configuration 1) or a single $c^{2}$ region (Configuration 2). The leftover spaces must be equal, so $a^{2} + b^{2} = c^{2}$ .

Step 4: Configuration 2: One Tilted Square + Four Triangles

Rearrange the same four triangles within the big square. Place them in the four corners, and the space remaining in the center forms a tilted blue square with side $c$ .

Total area = $c^{2} +$ (area of four triangles).

Step 5: Same Total, Same Four Triangles

Both configurations fill the same big square (area $(a + b)^{2}$ ) and use the same four identical triangles (total triangle area: $4 \times \frac{1}{2} ab = 2 ab$ ). The only difference is the remaining space!

Step 6: Conclusion: $a^{2} + b^{2} = c^{2}$

Subtracting the four triangles from both configurations: Configuration 1 gives $a^{2} + b^{2}$ , and Configuration 2 gives $c^{2}$ . Since these remaining spaces must be equal:

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

Since we only moved the triangles and didn't change the total area:

Big square - Four triangles = Remaining space

This remaining space must be equal in both configurations, proving:

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

Why This Proof Works

This proof is powerful because it:

Uses the conservation of area — moving shapes doesn't change their total area
Is visual and intuitive — you can "see" why the theorem is true
Requires no algebra — just geometric reasoning

Historical Context

This type of "dissection proof" has been used for centuries. Similar proofs appear in ancient Chinese mathematics (the Zhoubi Suanjing) and Indian mathematics (Bhāskara's proof). ∎