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: From the right-angled vertex \(A\), draw perpendicular \(AD\) to hypotenuse \(BC\), which intersects \(BC\) at point \(D\).
Proof: In the right-angled triangle \(ABC\), \(AD\) is the perpendicular to 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]