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 point A to the hypotenuse BC, which meets BC at point D.
Proof: Since AD is perpendicular to BC in the right-angled triangle ABC:
∆ABD ~ ∆CBA.
Thus, \(\frac{AB}{BC} = \frac{BD}{AB}\)
∴ \(AB^2 = BC \cdot BD\) …(i)
Similarly, ∆CAD ~ ∆CBA.
Thus, \(\frac{AC}{BC} = \frac{DC}{AC}\) …(ii)
Adding equations (i) and (ii),
\(AB^2 + AC^2 = BC \cdot BD + BC \cdot DC\)
= \(BC (BD + DC)\)
= \(BC \cdot BC = BC^2\)
∴ \(BC^2 = AB^2 + AC^2\) [Proved] Let me know if you need any refinements!