Right triangle midpoints – compute 4(AQ^2 + BP^2): In right-angled ΔABC (right angle at C), let P and Q be the midpoints of CA and CB respectively. Find the value of 4(AQ^2 + BP^2) in terms of AB.

Introduction / Context:
This is a coordinate-geometry application in a right triangle using midpoints and distance formulas, leading to a neat identity involving the hypotenuse.

Given Data / Assumptions:

• ΔABC right-angled at C.
• P is midpoint of CA; Q is midpoint of CB.
• We need 4(AQ^2 + BP^2).

Concept / Approach:
Place the triangle on coordinate axes for convenience: set C = (0, 0), A = (a, 0), B = (0, b). Then AB^2 = a^2 + b^2. Compute P and Q coordinates, then distances AQ and BP, and simplify.

Step-by-Step Solution:
Coordinates: A(a,0), B(0,b), C(0,0).P midpoint of CA ⇒ P(a/2, 0); Q midpoint of CB ⇒ Q(0, b/2).AQ^2 = (a − 0)^2 + (0 − b/2)^2 = a^2 + b^2/4.BP^2 = (0 − a/2)^2 + (b − 0)^2 = a^2/4 + b^2.Sum: AQ^2 + BP^2 = (a^2 + b^2/4) + (a^2/4 + b^2) = (5/4)(a^2 + b^2) = (5/4)AB^2.Multiply by 4: 4(AQ^2 + BP^2) = 5 AB^2.

Verification / Alternative check:
Try a = 3, b = 4 ⇒ AB = 5. Compute numerically to confirm the identity; it holds.

Why Other Options Are Wrong:
Other multiples of AB^2 do not match the coordinate computation.

Common Pitfalls:
Forgetting that Q is on CB (not AB), or mixing midpoint coordinates.

Final Answer:
5 AB^2