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: \(ABC\) 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-angled vertex \(A\) to the hypotenuse \(BC\), intersecting \(BC\) at point \(D\).
Proof: In the right-angled triangle \(ABC\), \(AD\) is perpendicular to the hypotenuse \(BC\).
\(∴\) \(∆ABD\) and \(∆CBA\) are similar.
Thus, \(\cfrac{AB}{BC} = \cfrac{BD}{AB}\)
\(∴ AB^2 = BC.BD ......(I)\)
Again, \(∆CAD\) and \(∆CBA\) are similar.
Thus, \(\cfrac{AC}{BC} = \cfrac{DC}{AC}……….(II)\)
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].