Q.3 (B)(4) Prove that, ‘In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of remaining two sides’.
Best answer
Answer: Proof of the Pythagoras theorem by similar triangles.
Given: In Δ ABC, ∠ABC = 90º. To prove: AC² = AB² + BC².
Construction: draw seg BD ⊥ hypotenuse AC, with D on AC.
Proof: In Δ ADB and Δ ABC, ∠ADB = ∠ABC = 90º (construction and given) and ∠A is common, so Δ ADB ~ Δ ABC by the AA test.
Corresponding sides are proportional: \(\dfrac{AD}{AB} = \dfrac{AB}{AC}\), so AB² = AD × AC ... (I).
Similarly, in Δ BDC and Δ ABC, ∠BDC = ∠ABC = 90º and ∠C is common, so Δ BDC ~ Δ ABC.
\(\dfrac{CD}{BC} = \dfrac{BC}{AC}\), so BC² = CD × AC ... (II).
Adding (I) and (II): AB² + BC² = AD × AC + CD × AC = AC(AD + CD) = AC × AC = AC².
Hence AC² = AB² + BC². Hence proved.
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