Right isosceles triangle relation:\nABC is an isosceles right triangle with ∠C = 90°. For any point D on hypotenuse AB, evaluate AD² + BD² in terms of CD.

Introduction / Context:
This identity leverages coordinate geometry in a right isosceles setup (legs equal, right angle at C). Placing the triangle conveniently on axes makes it easy to parametrize any point D on the hypotenuse AB and compute squared distances to A and B, then compare to CD².

Given Data / Assumptions:

• Right angle at C; AC = BC = t.
• A = (t, 0), B = (0, t), C = (0, 0).
• D lies on AB, which joins (t, 0) to (0, t).

Concept / Approach:
Parametrize D as A + u(B − A) = (t(1 − u), tu). Then compute AD², BD², CD². Simplify each in terms of u and t. Compare AD² + BD² with CD² to find the constant factor independent of u.

Step-by-Step Solution:

Verification / Alternative check:
Pick u = 1/2 (midpoint) to sanity check: then AD = BD, and AD² + BD² clearly doubles CD² by computation.

Why Other Options Are Wrong:
CD², 3CD², 4CD² contradict the derived identity valid for any D on AB.

Common Pitfalls:
Placing the triangle incorrectly or assuming D is a special point; the relation holds for any D on AB.

Final Answer:
2CD2