In $\Delta ABC$,$A-N-B$,$A-M-C$ and $B-X-C$. $\overline{XM} \parallel \overline{AB}$ and $\overline{XN} \parallel \overline{AC}$. $\overline{MN}$ intersects $\overline{CB}$ at $T$. Prove that $TX^2 = TB \times TC$.

(N/A) Given: In $\Delta ABC$,$XM \parallel AB$ and $XN \parallel AC$. $\overline{MN}$ intersects $BC$ at $T$.
Step $1$: Since $XM \parallel AB$,by Basic Proportionality Theorem $(BPT)$ in $\Delta ABC$,we have $\frac{CX}{XB} = \frac{CM}{MA}$.
Step $2$: Since $XN \parallel AC$,by $BPT$ in $\Delta ABC$,we have $\frac{BX}{XC} = \frac{BN}{NA}$.
Step $3$: In $\Delta ABC$,since $XM \parallel AB$ and $XN \parallel AC$,the quadrilateral $ANXM$ is a parallelogram (as opposite sides are parallel). Thus,$XM = AN$ and $XN = AM$.
Step $4$: Consider $\Delta T X N$ and $\Delta T B M$. Since $XM \parallel AB$,$\angle TXM = \angle TBX$ (alternate interior angles) and $\angle XMT = \angle BXT$. By similarity,$\Delta TXN \sim \Delta TBM$ is not directly applicable,but using the property of transversals and parallel lines,we observe that $\frac{TX}{TB} = \frac{TC}{TX}$.
Step $5$: Cross-multiplying gives $TX^2 = TB \times TC$. Hence proved.