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Proof of pythagoras theorem by using algebra and 3 other methods |
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Answer» Answer: Given: A ∆ XYZ in which ∠XYZ = 90°. To prove: XZ2 = XY2 + YZ2 Construction: Draw YO ⊥ XZ Proof: In ∆XOY and ∆XYZ, we have, ∠X = ∠X → common ∠XOY = ∠XYZ → each EQUAL to 90° Therefore, ∆ XOY ~ ∆ XYZ → by AA-similarity ⇒ XO/XY = XY/XZ ⇒ XO × XZ = XY2 ----------------- (i) In ∆YOZ and ∆XYZ, we have, ∠Z = ∠Z → common ∠YOZ = ∠XYZ → each equal to 90° Therefore, ∆ YOZ ~ ∆ XYZ → by AA-similarity ⇒ OZ/YZ = YZ/XZ ⇒ OZ × XZ = YZ2 ----------------- (ii) From (i) and (ii) we get, XO × XZ + OZ × XZ = (XY2 + YZ2) ⇒ (XO + OZ) × XZ = (XY2 + YZ2) ⇒ XZ × XZ = (XY2 + YZ2) ⇒ XZ 2 = (XY2 + YZ2) |
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