If in two triangles $ABC$ and $PQR$,$\frac{AB}{QR} = \frac{BC}{PR} = \frac{CA}{PQ}$,then

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AM = 4$ and $CM = 5$,find $AB$ and $BM$.

In the figure,two line segments $AC$ and $BD$ intersect each other at the point $P$ such that $PA = 6 \, cm$,$PB = 3 \, cm$,$PC = 2.5 \, cm$,$PD = 5 \, cm$,$\angle APB = 50^{\circ}$ and $\angle CDP = 30^{\circ}$. Then,$\angle PBA$ is equal to: (in $^{\circ}$)

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

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. Prove that $\frac{1}{BM^2} = \frac{1}{AB^2} + \frac{1}{BC^2}$.

In rectangle $ABCD$,$AB = 2.4$ and $BC = 3.2$. Then,$AC = \ldots$