Given: ABCD is a rectangle, and O is any point inside it.
To Prove: \( OA^2 + OC^2 = OB^2 + OD^2 \)
Construction: Draw a straight line through point O parallel to BC, which intersects AB and DC at points P and Q, respectively.
Proof: According to the construction, PQ || BC
∴ PQ ⊥ AB and PQ ⊥ DC
(∵ ∠B = 90° and ∠C = 90°)
∴ \( \triangle APO, \triangle BPO, \triangle CQO, \) and \( \triangle DQO \) are all right-angled triangles.
∴ \( OA^2 = AP^2 + OP^2 \)
\( OC^2 = CQ^2 + OQ^2 \)
\( OB^2 = BP^2 + OP^2 \)
\( OD^2 = DQ^2 + OQ^2 \)
∴ \( OA^2 + OC^2 = AP^2 + OP^2 + CQ^2 + OQ^2 \) ----(i)
But according to the construction, APQD and BPQC are each rectangles.
Thus, AP = DQ and CQ = BP.
From (i), we get:
\( OA^2 + OC^2 \)
\( = DQ^2 + OP^2 + BP^2 + OQ^2 \)
\( = (DQ^2 + OQ^2) + (BP^2 + OP^2) \)
\( = OD^2 + OB^2 \)
\( = OB^2 + OD^2 \) (Proved)