Point O inside rectangle ABCD is joined to all vertices. Which identity always holds?

Introduction / Context:The British Flag Theorem states that in a rectangle, for any point O (inside or outside), the sum of squares of distances from O to two opposite corners equals the sum for the other pair of opposite corners.

Given Data / Assumptions:

• ABCD is a rectangle; O is any point.
• We compare OA² + OC² with OB² + OD².

Concept / Approach:Using coordinates A(0,0), B(w,0), C(w,h), D(0,h), and O(x,y), compute squared distances. The algebra shows OA² + OC² = x² + y² + (w − x)² + (h − y)² equals OB² + OD² = (w − x)² + y² + x² + (h − y)².

Step-by-Step Solution:Expand both sums; cross terms cancel.Both totals reduce to w² + h² + 2(x² + y²) − 2(wx + hy) + 2(wx + hy) = w² + h² + 2(x² + y²).Hence OA² + OC² = OB² + OD².

Verification / Alternative check:A vector proof via parallelogram law in orthogonal axes arrives at the same identity.

Why Other Options Are Wrong:Minus signs or mixing equalities do not hold generally; option (d) is also true only if O is the rectangle’s centre, not for arbitrary O.

Common Pitfalls:Assuming it is true only for the rectangle’s centre; the theorem holds for any O.

Final Answer:OA2 + OC2 = OB2 + OD2