1. **Problem statement:** Prove the Pythagorean Theorem using the method of arranging four right-angled triangles inside a square. 2. **Given:** Four right-angled triangles with legs $a$, $b$ and hypotenuse $c$ are arranged inside a large square. 3. **Step 1: Calculate the area of the outer square.** The outer square has side length $c + c = 2c$ (since the hypotenuses form the sides), but from the problem description, the outer square side length is $a + b$. So, the area of the outer square is: $$\text{Area}_{outer} = (a + b)^2$$ 4. **Step 2: Calculate the area of the inner square.** The inner square is formed by the legs of the triangles and has side length: $$|a - b|$$ So, the area of the inner square is: $$\text{Area}_{inner} = (a - b)^2$$ 5. **Step 3: Calculate the area of one right-angled triangle.** Each triangle has area: $$\text{Area}_{triangle} = \frac{1}{2}ab$$ 6. **Step 4: Express the area of the outer square as the sum of the inner square and four triangles.** $$\text{Area}_{outer} = \text{Area}_{inner} + 4 \times \text{Area}_{triangle}$$ Substitute the values: $$ (a + b)^2 = (a - b)^2 + 4 \times \frac{1}{2}ab $$ Simplify the right side: $$ (a + b)^2 = (a - b)^2 + 2ab $$ 7. **Step 5: Expand both squares.** $$ a^2 + 2ab + b^2 = a^2 - 2ab + b^2 + 2ab $$ 8. **Step 6: Simplify the right side by canceling terms.** $$ a^2 + 2ab + b^2 = a^2 + b^2 $$ 9. **Step 7: Subtract $a^2 + b^2$ from both sides.** $$ \cancel{a^2} + 2ab + \cancel{b^2} - \cancel{a^2} - \cancel{b^2} = 0 $$ This simplifies to: $$ 2ab = 0 $$ This indicates an error in the previous step; let's re-express step 6 carefully. 10. **Re-express step 6 carefully:** Right side after expansion: $$ (a - b)^2 + 2ab = (a^2 - 2ab + b^2) + 2ab = a^2 - 2ab + b^2 + 2ab $$ Simplify: $$ a^2 + b^2 $$ So the equation is: $$ (a + b)^2 = a^2 + b^2 $$ But from step 3, the outer square side length is $a + b$, so: $$ (a + b)^2 = a^2 + b^2 + 2ab $$ Therefore, the correct equation is: $$ (a + b)^2 = (a - b)^2 + 4 \times \frac{1}{2}ab $$ Expanding: $$ a^2 + 2ab + b^2 = a^2 - 2ab + b^2 + 2ab $$ Simplify right side: $$ a^2 + b^2 $$ This shows: $$ a^2 + 2ab + b^2 = a^2 + b^2 $$ Subtract $a^2 + b^2$ from both sides: $$ 2ab = 0 $$ This is a contradiction, so the initial assumption about the outer square side length must be corrected. 11. **Correction:** The outer square side length is $c$, not $a + b$. Therefore: $$ \text{Area}_{outer} = c^2 $$ The inner square side length is $(a - b)$, so: $$ \text{Area}_{inner} = (a - b)^2 $$ The four triangles each have area $\frac{1}{2}ab$, so total area of triangles is: $$ 4 \times \frac{1}{2}ab = 2ab $$ 12. **Step 8: Write the area relation:** $$ c^2 = (a - b)^2 + 2ab $$ 13. **Step 9: Expand and simplify:** $$ c^2 = a^2 - 2ab + b^2 + 2ab $$ Simplify: $$ c^2 = a^2 + b^2 $$ 14. **Conclusion:** We have proved the Pythagorean Theorem: $$ c^2 = a^2 + b^2 $$ This completes the proof using the geometric arrangement of four right-angled triangles inside a square.