Question

In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then prove that the angle opposite to the first is a right angle.

Answer 1

Given: AC² = AB² + BC²

To prove: Angle opposite to the first side is a right angle.

Construction: Draw ΔPQR, where AB = PQ, BC = QR and ∠Q = 90°.

Proof: In ΔPQR,

PR² = PQ² + QR²   ...(By using Pythagoreas Theorem)

∵ AB = PQ, BC = QR   ...[From construction]

∴ PR² = AB² + BC²

Now, AC² = AB² + BC²   ...(Given)

∴ AC² = PR²

⇒ AC = PR

In ΔABC and ΔPQR,

AB = PQ   ...(By construction)

BC = QR   ...(By construction)

And AC = PR   ...(Proved above)

∴ ΔABC ≅ ΔPQR   ...(By SSS congruency rule)

So, ∠B = ∠Q   ...(By CPCT)

But ∠Q = 90°   ...(By construction)

Hence, ∠B = 90°

Hence Proved.