Given $\Delta ABC \sim \Delta XYZ$ for the correspondence $ABC \leftrightarrow XYZ$. If $AB : XY = 4 : 5$ and $YZ = 20$,find $BC$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BE}$ is an altitude to the hypotenuse $\overline{AC}$. Prove that $\frac{AB^2}{BC^2} = \frac{AE}{CE}$.

In the figure,altitudes $AD$ and $CE$ of $\Delta ABC$ intersect each other at the point $P$. Show that $\Delta AEP \sim \Delta ADB$.

In $\Delta ABC$ and $\Delta XYZ$,$m\angle A = m\angle X$,$m\angle B = m\angle Y$ and $\frac{AB}{XY} = \frac{2}{3}$. If $AC = 7.2$,then $XZ = \ldots$

$\Delta PQR \sim \Delta XYZ$ for the correspondence $PQR \leftrightarrow XZY$. If $\frac{PQ}{XZ} = \frac{7}{3}$ and the area of $\Delta XYZ$ is $72$,find the area of $\Delta PQR$.

For $\Delta ABC$ and $\Delta XYZ$,$\frac{AB}{XZ} = \frac{BC}{XY} = \frac{AC}{YZ}$. Then,the correspondence $ABC \leftrightarrow \ldots \ldots$ between them is a similarity.