**Pythagorean Theorem:** The area of the square constructed on the hypotenuse of a right-angled triangle is equal to the sum of the areas of the squares constructed on the other two sides. **Given:** \( \triangle ABC \) is a right-angled triangle with \( \angle A \) as the right angle. **To Prove:** \[ BC^2 = AB^2 + AC^2 \] **Construction:** Draw a perpendicular \( AD \) from the right-angled vertex \( A \) onto the hypotenuse \( BC \), intersecting \( BC \) at \( D \). **Proof:** Since \( AD \) is perpendicular to \( BC \), \( \triangle ABD \) and \( \triangle ABC \) are similar. Thus, \[ \frac{AB}{BC} = \frac{BD}{AB} \] \[ AB^2 = BC \cdot BD \quad \text{(i)} \] Similarly, \( \triangle CAD \) and \( \triangle ABC \) are similar. Thus, \[ \frac{AC}{BC} = \frac{DC}{AC} \quad \text{(ii)} \] Adding equations (i) and (ii), \[ AB^2 + AC^2 = BC \cdot BD + BC \cdot DC \] \[ = BC (BD + DC) \] \[ = BC \cdot BC = BC^2 \] Thus, \[ BC^2 = AB^2 + AC^2 \] (Proved).