In $\Delta XYZ$,$X-P-Y$,$X-Q-Z$ and $\overline{PQ} \parallel \overline{YZ}$. If $XP = 8$,$XY = 12$ and $XQ = 12$,find $QZ$.

If $\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR$,and given $AB + BC = 12$,$PQ + QR = 15$,and $AC = 8$,find the length of $PR$.

In rectangle $ABCD$,$AB = 9$ and $BC = 12$. Then,$BD = \ldots$

In the figure,if $\angle A = \angle C$,$AB = 6 \, cm$,$BP = 15 \, cm$,$AP = 12 \, cm$,and $CP = 4 \, cm$,then find the lengths of $PD$ and $CD$ (in $cm$).

In a triangle $PQR$,$N$ is a point on $PR$ such that $QN \perp PR$. If $PN \cdot NR = QN^2$,prove that $\angle PQR = 90^{\circ}$.

In $\square ABCD$,$P \in \overline{AB}$,$Q \in \overline{BC}$,$R \in \overline{CD}$,and $S \in \overline{AD}$ such that $\frac{AP}{AB} = \frac{CQ}{BC} = \frac{CR}{CD} = \frac{AS}{AD} = \frac{1}{3}$. Prove that $\square PQRS$ is a parallelogram.