In $\square ABCD$,$\overline{AB} \parallel \overline{CD}$ and $\overline{AC} \cap \overline{BD} = \{O\}$. If $\frac{OA}{OC} = \frac{1}{2}$ and $AB = 5$,find $CD$.

If $\Delta ABC \sim \Delta PQR$ under the correspondence $ABC \leftrightarrow QPR$,and given $m \angle A + m \angle B = 130^{\circ}$ and $m \angle B + m \angle C = 125^{\circ}$,find $m \angle Q$. (in $^{\circ}$)

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. Then,$\angle BAM \cong \ldots$

If $\Delta ABC \sim \Delta PQR$ under the correspondence $ABC \leftrightarrow RPQ$,and $m\angle A = 35^{\circ}$ and $m\angle B = 50^{\circ}$,find $m\angle Q$. (in $^{\circ}$)

In the figure,$ABD$ is a triangle right-angled at $A$ and $AC \perp BD$. Show that $AB^2 = BC \cdot BD$.

If in triangles $ABC$ and $DEF$,$\frac{AB}{DE} = \frac{BC}{FD}$,then they will be similar,when