In ΔABC, ∠C = 90°, point P and Q are on the sides AC and BC, respectively, such that AP : PC = BQ : QC = 1 : 2. Then, A Q^2 + B P^2 A B^2 is equal to:

Answer: Option C

Let the side of BC & CA are 3x and 3y
QC = 2x
QB = x
PC = 2y
AP = y
In triangle AQC
AQ² = (3y)² + (2x)²
= 9y² + 4x²
BP² = (3x)² + (2y)²
= 9x² + 4y²
AB² = (3x)² + (3y)²
= 9x² + 9y²
= 9(x² + y²)
$$\text{Now, }\frac{\text{A}{\text{Q}}^2+\text{B}{\text{P}}^2}{\text{A}{\text{B}}^2}=\frac{13\left(x^2+y^2\right)}{9\left(x^2+y^2\right)}=\frac{13}{9}$$