Triangle ΔABC is an isosceles right angled triangle with ∠C = 90°. Point D is any point on the hypotenuse AB. What is the value of AD^2 + BD^2 in terms of CD^2?

Introduction / Context:
This is a geometry identity question involving a special triangle, namely an isosceles right angled triangle, and a point on its hypotenuse. It explores a lesser known but elegant relationship between the squares of distances from that point to the endpoints of the hypotenuse and the distance from the point to the right angle vertex. Such identities often arise in coordinate geometry or vector geometry and are useful in advanced problem solving.

Given Data / Assumptions:

• ΔABC is an isosceles right angled triangle.
• ∠C = 90°, so C is the right angle.
• CA = CB, so the legs adjacent to the right angle are equal.
• AB is the hypotenuse.
• D is any point on AB.
• We are asked to express AD^2 + BD^2 in terms of CD^2.

Concept / Approach:
The cleanest way to analyse this is to place the triangle in a coordinate system. Because the triangle is isosceles and right angled at C, we can put C at the origin, A on the x axis, and B on the y axis with equal leg lengths. Then any point D on the hypotenuse has parametric coordinates depending on a single parameter. By computing squared distances AD^2, BD^2 and CD^2 in terms of this parameter, we can look for a pattern that links AD^2 + BD^2 to CD^2. For an isosceles right triangle, a very neat simplification occurs.

Step-by-Step Solution:
Place C at (0, 0), A at (a, 0) and B at (0, a) so CA = CB = a. Then AB is the segment from (a, 0) to (0, a). Let D be any point on AB, for example D = (t, a - t) for some real t between 0 and a. Compute AD^2 = (t - a)^2 + (a - t)^2 = 2(t - a)^2. Compute BD^2 = t^2 + (a - t - a)^2 = t^2 + (-t)^2 = 2t^2. So AD^2 + BD^2 = 2(t - a)^2 + 2t^2 = 4t^2 - 4at + 2a^2. Now compute CD^2 = t^2 + (a - t)^2 = 2t^2 - 2at + a^2. Notice that AD^2 + BD^2 = 2 * CD^2.

Verification / Alternative check:
You can plug in specific values for a and t to check the relationship numerically. For instance, let a = 1 and t = 0.2. Compute CD^2 and AD^2 + BD^2 numerically and you will find that AD^2 + BD^2 is very close to twice CD^2, accounting for rounding. Since the derivation above is algebraic and general, it holds for any valid position of D on AB, confirming that the identity AD^2 + BD^2 = 2CD^2 is true for this triangle.

Why Other Options Are Wrong:
CD^2: This would mean AD^2 + BD^2 equals CD^2, which is not supported by the coordinate derivation. 3CD^2 and 4CD^2: These factors overstate the combined squared distances and do not match any consistent algebraic identity for the situation described. AD^2: This does not involve CD at all and is unrelated to the requested expression.

Common Pitfalls:
Assuming D is a special point such as the midpoint of AB rather than any point on AB. Trying to use Pythagoras directly without setting up a coordinate or vector framework. Not recognising that symmetry in an isosceles right triangle often leads to simple numeric factors like 2.

Final Answer:
For any point D on AB, we have AD^2 + BD^2 = 2CD^2