1. **State the problem:** Prove the Pythagorean Theorem using the method of arranging four right-angled triangles around a square of side length $c$. 2. **Formula and setup:** The outer large square has side length $(a+b)$, so its area is $$(a+b)^2.$$ Inside it, there is a smaller square with side length $c$, so its area is $$c^2.$$ Four right-angled triangles each have legs $a$ and $b$, so the area of one triangle is $$\frac{1}{2}ab.$$ 3. **Relationship between areas:** The large square's area equals the area of the smaller square plus the area of the four triangles: $$ (a+b)^2 = c^2 + 4 \times \frac{1}{2}ab $$ 4. **Simplify the right side:** $$ (a+b)^2 = c^2 + 2ab $$ 5. **Expand the left side:** $$ (a+b)^2 = a^2 + 2ab + b^2 $$ 6. **Set the two expressions equal:** $$ a^2 + 2ab + b^2 = c^2 + 2ab $$ 7. **Subtract $2ab$ from both sides:** $$ a^2 + \cancel{2ab} + b^2 = c^2 + \cancel{2ab} $$ 8. **Final result:** $$ a^2 + b^2 = c^2 $$ This completes the proof of the Pythagorean Theorem using the given geometric arrangement.