Prove that in a right-triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Given: A right angled ∆ABC, right angled at B

To Prove: AC² = AB² + BC²

Construction: Draw perpendicular BD onto the side AC i.e. BD ⊥ AC

Proof: Since BD ⊥ AC

Using Theorem,

If a perpendicular is drawn from the vertex of a right angle of a right angled triangle to the hypotenuse, then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

We have

ΔADB ∼ ΔABC. (by AA similarity)

Since side of similar triangles are in the same ratio,

(In similar Triangles corresponding sides are proportional)

⇒ AB² = AD × AC …(1)

Also, ΔBDC ∼ ΔABC

(In similar Triangles corresponding sides are proportional)

⇒ BC² = CD × AC …(2)

Adding the equations (1) and (2) we get,

AB² + BC² = AD × AC + CD × AC

⇒ AB² + BC² = AC (AD + CD)

(From the figure AD + CD = AC)

⇒ AB² + BC² = AC × AC

Therefore, AC² = AB² + BC²

This theorem is known as Pythagoras theorem