**Pythagoras' Theorem:** The area of the square drawn on the hypotenuse of a right-angled triangle is equal to the sum of the areas of the squares drawn on the other two sides. **Given:** 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 point D. **Proof:** Since AD is perpendicular to BC in the right-angled triangle ABC, ∴ \(\triangle\)ABD and \(\triangle\)CBA are similar. Thus, \( \frac{AB}{BC} = \frac{BD}{AB} \) ∴ \( AB^2 = BC.BD \) ....(i) Similarly, \(\triangle\)CAD and \(\triangle\)CBA are similar. Thus, \( \frac{AC}{BC} = \frac{DC}{AC} \) ....(ii) Adding (i) and (ii), we get: \( AB^2 + AC^2 = BC.BD + BC.DC \) \( = BC (BD + DC) \) \( = BC.BC = BC^2 \) ∴ \(BC^2 = AB^2 + AC^2\) [Proved]