Proof 2 of Pythagorean Theorem: The Puzzle Shift

A visual dissection proof showing two ways to fill an $(a+b) \times (a+b)$ square. Both use four identical right triangles (legs $a$ and $b$). The remaining space is $a^2 + b^2$ in one arrangement, $c^2$ in the other. Therefore $a^2 + b^2 = c^2$.

Prerequisites: pythagorean-theorem · Difficulty: beginner

This proof treats the Pythagorean theorem like a puzzle where you move pieces around without adding or taking anything away. It's a visual, intuitive proof that shows how rearranging shapes preserves area.

The Key Idea

Start with a large square with side length $(a + b)$. We'll fill this square in two different ways, both using four identical right triangles (each with legs $a$ and $b$, and hypotenuse $c$).

If we can show two different arrangements — both using the same four triangles — then whatever space remains after placing the triangles must be equal in both cases.

• Configuration 1: A green $a \times a$ square + a red $b \times b$ square + four triangles
• Configuration 2: A tilted blue $c \times c$ square + the same four triangles

Since the total area is the same and the triangles are identical, the remaining spaces must be equal: $a^2 + b^2 = c^2$.

The Proof

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: $4 \times \frac{1}{2}ab = 2ab$). 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
    

Conclusion

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

\text{Big square} - \text{Four triangles} = \text{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).