I have drawn a triangle PQR whose ∠Q is right angle. If S is any point on QR, let us prove that, PS² + QR² = PR² + QS²

Given: ∠Q is a right angle.

To prove: PS² + QR²= PR² + QS²

Proof:

It can be seen both the triangles ΔPQS and ΔPQR are right angled triangles.

Applying Pythagoras Theorem to ΔPQS gives,

PS² = PQ² + QS² [∵ H² = P² + B²]

⇒ PQ² = PS² - QS² ……………… (1)

Applying Pythagoras theorem to ΔPQR gives,

PR² = PQ² + QR² [∵ H² = P² + B²]

Substituting PQ² from (1) gives,

PR² = (PS² - QS²) + QR²

⇒ PR² = PS² - QS² + QR²

⇒ PR² + QS² = PS² + QR²

⇒ PS² + QR²= PR² + QS²

Hence proved.