Q.If the area of the square drawn on one side of any triangle is equal to the sum of the areas of the squares drawn on the other two sides, then prove that the angle opposite to the first side is a right angle.

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.
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