In trapezium $ABCD$,$\overline{AB} \parallel \overline{CD}$ and $\overline{AC} \cap \overline{BD} = \{P\}$. If $PA = 10$,$PC = 15$ and $PD = 12$,then $BD = \ldots$

In $\Delta ABC$ and $\Delta PQR$,if $\frac{AB}{PQ} = \frac{BC}{PR} = \frac{CA}{QR}$,then the correspondence $ABC \leftrightarrow \dots$ is a similarity.

If $\Delta ABC \sim \Delta XYZ$ under the correspondence $ABC \leftrightarrow XZY$,and the perimeter of $\Delta ABC$ is $45$,the perimeter of $\Delta XYZ$ is $30$,and $AB = 21$,find the length of $XZ$.

$D$ is a point on the side $BC$ of a triangle $ABC$ such that $\angle ADC = \angle BAC$. Show that $CA^2 = CB \cdot CD$.

If $\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR$,$2 AB = PQ$,and $BC = 10$,then $QR = \dots$

$\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR$. $\overline{AD}$ and $\overline{PM}$ are medians of these triangles. Prove that $AB \times PM = PQ \times AD$.