In $\Delta ABC$,$m\angle B = 90^{\circ}$. $D$ is the midpoint of $\overline{BC}$ and $F$ is the midpoint of $\overline{AB}$. Prove that $AD^{2} + CF^{2} = \frac{5}{4} AC^{2}$.

(N/A) $1$. In $\Delta ABC$,by the Pythagorean theorem,$AC^{2} = AB^{2} + BC^{2}$.
$2$. In right-angled $\Delta ABD$,$AD^{2} = AB^{2} + BD^{2}$. Since $D$ is the midpoint of $BC$,$BD = \frac{1}{2} BC$. Thus,$AD^{2} = AB^{2} + (\frac{1}{2} BC)^{2} = AB^{2} + \frac{1}{4} BC^{2}$.
$3$. In right-angled $\Delta CBF$,$CF^{2} = BC^{2} + BF^{2}$. Since $F$ is the midpoint of $AB$,$BF = \frac{1}{2} AB$. Thus,$CF^{2} = BC^{2} + (\frac{1}{2} AB)^{2} = BC^{2} + \frac{1}{4} AB^{2}$.
$4$. Adding the two equations: $AD^{2} + CF^{2} = (AB^{2} + \frac{1}{4} BC^{2}) + (BC^{2} + \frac{1}{4} AB^{2})$.
$5$. Simplifying: $AD^{2} + CF^{2} = (1 + \frac{1}{4}) AB^{2} + (1 + \frac{1}{4}) BC^{2} = \frac{5}{4} AB^{2} + \frac{5}{4} BC^{2}$.
$6$. Factoring out $\frac{5}{4}$: $AD^{2} + CF^{2} = \frac{5}{4} (AB^{2} + BC^{2})$.
$7$. Since $AB^{2} + BC^{2} = AC^{2}$,we get $AD^{2} + CF^{2} = \frac{5}{4} AC^{2}$.