Pythagorean 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: From the right angle at point A, draw a perpendicular AD to the hypotenuse BC, intersecting it at point D.
Proof: In the right-angled triangle ABC, a perpendicular AD is drawn to the hypotenuse BC.
∴ ∆ABD and ∆CBA are similar.
So, \(\frac{AB}{BC} = \frac{BD}{AB}\)
∴ \(AB^2 = BC \cdot BD \quad ...(i)\)
Again, ∆CAD and ∆CBA are similar.
So, \(\frac{AC}{BC} = \frac{DC}{AC} \quad ...(ii)\)
Adding equations (i) and (ii), we get:
\(AB^2 + AC^2 = BC \cdot BD + BC \cdot DC\)
\(= BC (BD + DC)\)
\(= BC \cdot BC = BC^2\)
∴ \(BC^2 = AB^2 + AC^2\) [Proved]