Let us assume that in triangle △ABC, the area of the square drawn on side AB is equal to the sum of the areas of the squares drawn on sides BC and AC. That is, \[ AB^2 = AC^2 + BC^2 \] We need to prove that ∠ACB is a right angle. **Construction:** Draw a line segment FE equal in length to BC. From point F, draw a perpendicular line and mark a point D on it such that FD = AC. Now join points D and E. **Proof:** Given: \[ AB^2 = BC^2 + AC^2 \Rightarrow AB^2 = EF^2 + DF^2 \quad [\text{Since } EF = BC \text{ and } DF = AC \text{ by construction}] \Rightarrow AB^2 = DE^2 \quad [\text{Since } ∠DFE = 90°, by construction] \Rightarrow AB = DE \] Now, in triangles △ABC and △DEF: - AB = DE - BC = EF - AC = DF So, \[ △ABC \cong △DEF \quad \text{[By SSS congruence criterion]} \Rightarrow ∠ACB = ∠DEF = 90° \quad [\text{Since } DF ⊥ EF by construction] \] Therefore, ∠ACB is a right angle — proven.