In $\Delta ABC$,$D$,$E$,and $F$ are the midpoints of $\overline{AB}$,$\overline{BC}$,and $\overline{CA}$ respectively. Then the correspondence $DEF \leftrightarrow \dots$ is a similarity.

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

If $\Delta ABC \sim \Delta XYZ$ under the correspondence $ABC \leftrightarrow XZY$,then $BC^2 : YZ^2 = \ldots \ldots \ldots$ (Wait,the question asks for $ABC : XYZ = BC^2 : \ldots$ which refers to the ratio of areas). Given $\Delta ABC \sim \Delta XZY$,the ratio of their areas is equal to the ratio of the squares of their corresponding sides. Thus,$\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta XZY)} = \frac{BC^2}{ZY^2}$.

$\Delta PQR \sim \Delta XYZ$ for the correspondence $PQR \leftrightarrow XZY$. If $PQ = 8$,$XZ = 12$ and the area of $\Delta PQR$ is $24$,find the area of $\Delta XYZ$.

$\Delta ABC \sim \Delta XYZ$ for the correspondence $ABC \leftrightarrow XYZ$. If $AB + BC = 7$,$XY + YZ = 10.5$,and $AC = 5$,then $XZ = \ldots$

If $\Delta ABC \sim \Delta PQR$ under the correspondence $ABC \leftrightarrow PQR$. If $AB = 12$,$\text{Area}(\Delta ABC) = 36$,and $\text{Area}(\Delta PQR) = 64$,then $PQ = \ldots$