In a quadrilateral PQRS, ∠Q = ∠S = 90° then prove that 2PR² - QR² = PQ² + PS² + SR².

In quadrilateral PQRS, since ∠Q = ∠S = 90°, triangles PQR and PSR are right-angled triangles.

According to Pythagoras theorem,

In a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

⇒ Hypotenuse² = Base² + Height²

In triangle PQR,

⇒ PR² = PQ² + QR² ………………….(1)

In triangle PSR,

⇒ PR² = PS² + SR² ………………(2)

Adding eq (1) and (2):

⇒ PR² + PR² = PQ² + QR² + PS² + SR²

⇒ 2PR² = PQ² + PS² + SR² + QR²

⇒ 2PR² - QR² = PQ² + PS² + SR²

Hence, proved that 2PR² - QR² = PQ² + PS² + SR².