Given: a ∆ xyz in which ∠xyz = 90°

Theorem.(Pythagorean theorem.) The square of the hypotenuse of a right triangle is equal to the sum of the squares of its legs.

Given: A ∆ XYZ in which ∠XYZ = 90°.

XY = a. XZ = c. YZ = b
To prove: XZ^2 = XY^2 + YZ^2
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 = XY^2
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 = YZ^2

XO × XZ + OZ × XZ = (XY^2 + YZ^2)

⇒ (XO + OZ) × XZ = (XY^2+ YZ^2)
⇒ XZ × XZ = (XY^2 + YZ^2)
⇒ XZ^2 = XY^2 + YZ^2. ⇒. c^2=a^2+b^2