In $\Delta ABC$,$\Delta PQR$,and $\Delta XYZ$,the correspondences $ABC \leftrightarrow QPR$ and $PQR \leftrightarrow YZX$ are similarities. Then,the correspondence $ABC \leftrightarrow \dots$ is a similarity.

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

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 $^{\circ}$)

$\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR$. If $AB : PQ = 3 : 4$ and the perimeter of $\Delta PQR$ is $24$,find the perimeter of $\Delta ABC$.

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,$\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta XZY)} = \frac{BC^2}{ZY^2}$.

If the areas of two similar triangles are equal,prove that they are congruent.