Let us consider a triangle ABC, in which

AC² = BC² + AB² …[1]

To Prove: Angle opposite to the first side i.e. AC is right angle or

∠ ABC = 90°

Construction:

Let us draw another right-angled triangle PQR right-angled at B, with

AB = PQ

BC = QR

Now, By Pythagoras theorem, In ΔPQR

PR² = QR² + PQ²

But QR = BC and PQ = AB

⇒ PR² = BC² + AB²

But From [1] we have,

AC² = PR²

⇒ AC = PR

In ΔABC and ΔPQR

AB = PQ [Assumed]

BC = QR [Assumed]

AC = PR [Proved above]

⇒ ΔABC ≅ ΔPQR [By Side-Side-Side Criterion]

⇒ ∠ABC = ∠PQR [Corresponding parts of congruent triangles are equal]

But, ∠PQR = 90°

⇒ ∠ABC = 90°

Hence, Proved !