Pythagorean Theorem: The area of the square drawn on the hypotenuse of any 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 where \(∠A\) is the right angle. To Prove: \( BC^2 = AB^2 + AC^2 \) Construction: A perpendicular \( AD \) is drawn from the right-angle vertex \( A \) to the hypotenuse \( BC \), intersecting it at point \( D \). Proof: In the right-angled triangle \( ABC \), \( AD \) is perpendicular to the hypotenuse \( BC \). \(∴ ∆ABD \) and \( ∆CBA \) are similar. Thus, \[ \cfrac{AB}{BC} = \cfrac{BD}{AB} \] \[ ∴ AB^2 = BC.BD \quad ......(I) \] Similarly, \( ∆CAD \) and \( ∆CBA \) are similar. Thus, \[ \cfrac{AC}{BC} = \cfrac{DC}{AC} \quad ......(II) \] Adding equations (I) and (II), \[ AB^2 + AC^2 = BC.BD + BC.DC \] \[ = BC (BD + DC) \] \[ = BC.BC = BC^2 \] \(∴ BC^2 = AB^2 + AC^2 \) [Proved]