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, $$\frac{{{\text{A}}{{\text{Q}}^2} + {\text{B}}{{\text{P}}^2}}}{{{\text{A}}{{\text{B}}^2}}}$$   is equal to:

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)
$${\text{Now, }}\frac{{{\text{A}}{{\text{Q}}^2} + {\text{B}}{{\text{P}}^2}}}{{{\text{A}}{{\text{B}}^2}}} = \frac{{13\left( {{x^2} + {y^2}} \right)}}{{9\left( {{x^2} + {y^2}} \right)}} = \frac{{13}}{9}$$