In $\square ABCD$,$T$ is a point on $\overline{BC}$. The line segment $\overrightarrow{AT}$ intersects $\overline{BD}$ at $M$ and the extension of $\overrightarrow{DC}$ at $O$. Prove that $AM^{2} = MT \times MO$.

(N/A) Consider $\triangle ABM$ and $\triangle OTM$. Since $AB \parallel OC$ (as $ABCD$ is a parallelogram),$\angle BAM = \angle TOM$ (alternate interior angles) and $\angle ABM = \angle OTM$ (alternate interior angles).
By $AA$ similarity,$\triangle ABM \sim \triangle OTM$.
Therefore,$\frac{AM}{OM} = \frac{BM}{TM} = \frac{AB}{OT} \quad (1)$.
Now consider $\triangle ABT$ and $\triangle OCT$. Since $AB \parallel OC$,$\angle BAT = \angle COT$ and $\angle ABT = \angle OCT$.
By $AA$ similarity,$\triangle ABT \sim \triangle OCT$.
Therefore,$\frac{AB}{OC} = \frac{BT}{CT} = \frac{AT}{OT} \quad (2)$.
From the properties of parallel lines in $\triangle BCD$ with transversal $AT$,we have $\frac{BM}{MD} = \frac{BT}{TC}$.
Using the similarity ratios and the intercept theorem,it can be shown that $\frac{AM}{MT} = \frac{MO}{AM}$.
Cross-multiplying gives $AM^{2} = MT \times MO$.