In parallelogram $ABCD$,$A-P-B$ and $AP = \frac{2}{3} AB$. $\overline{DP}$ intersects $\overline{AC}$ at $Q$. Find the ratio of the areas of $\Delta APQ$ and $\Delta CDQ$.

If $\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR$,and the perimeter of $\Delta ABC$ is $48$ while the perimeter of $\Delta PQR$ is $60$,find the ratios $\frac{AB}{PQ}$ and $\frac{AB + BC}{PQ + QR}$.

If $AD$ and $PM$ are medians of triangles $ABC$ and $PQR,$ respectively,where $\Delta ABC \sim \Delta PQR,$ prove that $\frac{AB}{PQ} = \frac{AD}{PM}.$

In $\Delta PQR$,$m \angle Q = 90^{\circ}$ and $\overline{QD}$ is an altitude. If $PD = 9 DR$,then $PQ = \ldots \ldots \ldots \times QR$.

Sides $AB$ and $BC$ and median $AD$ of a triangle $ABC$ are respectively proportional to sides $PQ$ and $QR$ and median $PM$ of $\Delta PQR$ (see Figure). Show that $\Delta ABC \sim \Delta PQR$.

In the figure,$D$ is a point on the hypotenuse $AC$ of $\Delta ABC,$ such that $BD \perp AC,$ $DM \perp BC,$ and $DN \perp AB.$ Prove that $DM^{2} = DN \cdot MC.$