Prove the Pythagoras theorem

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) 

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)