Prove that in a right-angled triangle, the area of the square constructed on the hypotenuse is equal to the sum of the areas of the squares constructed on the other two sides.

Q.Prove that in a right-angled triangle, the area of the square constructed on the hypotenuse is equal to the sum of the areas of the squares constructed on the other two sides.

Given: ABC is a right-angled triangle where ∠A is the right angle. To Prove: \(BC^2 = AB^2 + AC^2\) Construction: From the right-angle vertex A, draw a perpendicular AD to the hypotenuse BC, intersecting it at point D. **Proof:** In the right-angled triangle ABC, AD is perpendicular to the hypotenuse BC. ∴ Triangles ∆ABD and ∆CBA are similar. So, \[ \frac{AB}{BC} = \frac{BD}{AB} \Rightarrow AB^2 = BC \cdot BD \quad \text{...(i)} \] Again, triangles ∆CAD and ∆CBA are similar. So, \[ \frac{AC}{BC} = \frac{DC}{AC} \Rightarrow AC^2 = BC \cdot DC \quad \text{...(ii)} \] Adding equations (i) and (ii): \[ AB^2 + AC^2 = BC \cdot BD + BC \cdot DC = BC(BD + DC) = BC \cdot BC = BC^2 \] ∴ \(BC^2 = AB^2 + AC^2\) — Proved.