In $\Delta ABC$,$D \in \overline{AB}$,$E \in \overline{AC}$ and $\overline{DE} \parallel \overline{BC}$. $F$ is a point on $\overline{AD}$ such that $\overline{EF} \parallel \overline{CD}$. Prove that $AD^2 = AB \times AF$.

(N/A) Given: In $\Delta ABC$,$\overline{DE} \parallel \overline{BC}$ and $\overline{EF} \parallel \overline{CD}$.
Step $1$: In $\Delta ABC$,since $\overline{DE} \parallel \overline{BC}$,by the Basic Proportionality Theorem $(BPT)$,we have $\frac{AD}{AB} = \frac{AE}{AC}$.
Step $2$: In $\Delta ADC$,since $\overline{EF} \parallel \overline{CD}$,by the Basic Proportionality Theorem $(BPT)$,we have $\frac{AF}{AD} = \frac{AE}{AC}$.
Step $3$: Comparing the two equations from Step $1$ and Step $2$,we get $\frac{AD}{AB} = \frac{AF}{AD}$.
Step $4$: By cross-multiplying,we obtain $AD^2 = AB \times AF$. Hence proved.