1. **State the problem:** Prove the Pythagorean Theorem using the method of arranging four right-angled triangles inside a square. 2. **Formula and setup:** We have a large square with side length $(a+b)$ formed by arranging four right-angled triangles with sides $a$, $b$, and hypotenuse $c$, and a smaller inner square with side length $c$. 3. **Calculate the area of the outer square:** $$\text{Area}_{\text{outer}} = (a+b)^2$$ 4. **Calculate the area of the inner square:** $$\text{Area}_{\text{inner}} = c^2$$ 5. **Calculate the area of one right-angled triangle:** $$\text{Area}_{\triangle} = \frac{1}{2}ab$$ 6. **Express the total area of the outer square as the sum of the inner square and four triangles:** $$\text{Area}_{\text{outer}} = \text{Area}_{\text{inner}} + 4 \times \text{Area}_{\triangle}$$ 7. **Substitute the values:** $$ (a+b)^2 = c^2 + 4 \times \frac{1}{2}ab $$ 8. **Simplify the right side:** $$ (a+b)^2 = c^2 + 2ab $$ 9. **Expand the left side:** $$ a^2 + 2ab + b^2 = c^2 + 2ab $$ 10. **Subtract $2ab$ from both sides:** $$ a^2 + \cancel{2ab} + b^2 = c^2 + \cancel{2ab} $$ $$ a^2 + b^2 = c^2 $$ 11. **Conclusion:** This proves the Pythagorean Theorem: the square of the hypotenuse $c$ equals the sum of the squares of the other two sides $a$ and $b$.