PORS is a rectangle, and A is any point inside the rectangle. Prove that: \[ AP^2 + AR^2 = AQ^2 + AS^2 \]

Q.PORS is a rectangle, and A is any point inside the rectangle. Prove that: \[ AP^2 + AR^2 = AQ^2 + AS^2 \]

Given: A is any point inside the rectangle PQRS. To Prove: \[ AP^2 + AR^2 = AQ^2 + AS^2 \] Construction: Draw a straight line through point A parallel to QR, which intersects sides AB and DC at points M and N respectively. Proof: From the construction, QR ∥ MN ∴ MN ⊥ PQ and MN ⊥ RS (∵ ∠Q = 90° and ∠R = 90° in the rectangle) Therefore, triangles ∆PMA, ∆QMA, ∆RNA, and ∆SNA are all right-angled triangles. Using the Pythagorean theorem: \[ AP^2 = PM^2 + AM^2 AR^2 = NR^2 + AN^2 AQ^2 = MQ^2 + AM^2 AS^2 = SN^2 + AN^2 \] So, \[ AP^2 + AR^2 = PM^2 + AM^2 + NR^2 + AN^2 = SN^2 + AN^2 + MQ^2 + AM^2 = (SN^2 + AN^2) + (MQ^2 + AM^2) = AS^2 + AQ^2 = AQ^2 + AS^2 \] proved: