It is given that $\triangle ABC \sim \triangle DFE$,$\angle A = 30^{\circ}$,$\angle C = 50^{\circ}$,$AB = 5 \, cm$,$AC = 8 \, cm$ and $DF = 7.5 \, cm$. Then,the following is true:

In $\Delta ABC$,the bisector of $\angle B$ intersects $\overline{AC}$ at $D$. If $\frac{AD}{DC} = \frac{3}{4}$ and $AB = 7.5$,then $BC = \ldots$

$\Delta ABC \sim \Delta XYZ$ for the correspondence $ABC \leftrightarrow XYZ$. $\overline{AD} \perp \overline{BC}$,$D \in \overline{BC}$ and $\overline{XM} \perp \overline{YZ}$,$M \in \overline{YZ}$. Prove that $\frac{AD}{XM} = \frac{BC}{YZ}$.

In $\Delta PQR$,$m\angle Q = 90^{\circ}$ and $\overline{QM}$ is an altitude. If $PM = x$ and $RM = y$,find the lengths of $\overline{PQ}$,$\overline{QR}$,$\overline{PR}$,and $\overline{QM}$ in terms of $x$ and $y$.

In $\Delta ABC$,$B-D-C$. The bisectors of $\angle ADB$ and $\angle ADC$ intersect $\overline{AB}$ and $\overline{AC}$ at $P$ and $Q$ respectively. Prove that $AP \times AQ \times BD \times DC = AD^2 \times PB \times QC$. From this,prove that $D$ is the midpoint of $\overline{BC}$ if $\overleftrightarrow{PQ} \parallel \overleftrightarrow{BC}$.

$\Delta ABC$ is an acute-angled triangle and $\overline{AM}$ is an altitude. Prove that $AC^{2} = AB^{2} + BC^{2} - 2 \cdot BC \cdot BM$.