In parallelogram ABCD with diagonals AC and BD, which identity is true?

Introduction / Context:
There is a classical relation in any parallelogram linking the sums of squares of its sides with the sums of squares of its diagonals. This identity is frequently used in vector geometry and Euclidean proofs.

Given Data / Assumptions:

• ABCD is a parallelogram (AB ∥ CD, BC ∥ AD; opposite sides equal).
• Diagonals are AC and BD.

Concept / Approach:

• Vector method: Let vectors a = AB and b = AD. Then AC = a + b and BD = a − b.
• Use norm properties: |a + b|^2 + |a − b|^2 = 2|a|^2 + 2|b|^2.
• Translate back to side lengths: AB = |a|, BC = |b|, CD = |a|, DA = |b|.

Step-by-Step Solution:

Verification / Alternative check:
Special case rectangle: AC^2 + BD^2 = 2(a^2 + b^2); LHS = 2a^2 + 2b^2 — identity holds.

Why Other Options Are Wrong:

• Options (a), (c), (d) misstate the sign/factors; only (b) matches the parallelogram identity.
• None of these: Not applicable; (b) is correct.

Common Pitfalls:

• Forgetting that AB = CD and BC = AD when summing squares.

Final Answer:
AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + BD^2