Question

proof of pythagoras theorem by using algebra and 3 other methods​

Answer 1

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)