Pythagoras' Theorem: In any right-angled triangle, the area of the square drawn on the hypotenuse 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: Draw a perpendicular \(AD\) from the right-angled point \(A\) to the hypotenuse \(BC\), which intersects \(BC\) at point \(D\).
Proof: The perpendicular \(AD\) divides \(∆ABC\) into two similar triangles \(∆ABD\) and \(∆CBA\).
Thus, \( \cfrac{AB}{BC} = \cfrac{BD}{AB} \).
\(∴ AB^2 = BC.BD \) ......(I)
Similarly, \(∆CAD\) and \(∆CBA\) are also similar.
So, \( \cfrac{AC}{BC} = \cfrac{DC}{AC} \) ……….(II)
Adding equations \((I)\) 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).