In $\Delta ABC$,$AB = 6\sqrt{3} \text{ cm}$,$AC = 12 \text{ cm}$,and $BC = 6 \text{ cm}$. The angle $B$ is (in $^o$):

Sides $AB$ and $AC$ and median $AD$ of a triangle $ABC$ are respectively proportional to sides $PQ$ and $PR$ and median $PM$ of another triangle $PQR$. Show that $\Delta ABC \sim \Delta PQR$.

$CD$ and $GH$ are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that $D$ and $H$ lie on sides $AB$ and $FE$ of $\Delta ABC$ and $\Delta EFG$ respectively. If $\Delta ABC \sim \Delta FEG,$ show that:
$(i) \frac{CD}{GH} = \frac{AC}{FG}$
$(ii) \Delta DCB \sim \Delta HGE$
$(iii) \Delta DCA \sim \Delta HGF$

In the figure,$ABD$ is a triangle right-angled at $A$ and $AC \perp BD$. Show that $AB^2 = BC \cdot BD$.

In Figure,$CM$ and $RN$ are respectively the medians of $\Delta ABC$ and $\Delta PQR$. If $\Delta ABC \sim \Delta PQR$,prove that:
$(i)$ $\Delta AMC \sim \Delta PNR$
$(ii)$ $\frac{CM}{RN} = \frac{AB}{PQ}$
$(iii)$ $\Delta CMB \sim \Delta RNQ$

In the figure,$D$ is a point on the hypotenuse $AC$ of $\Delta ABC,$ such that $BD \perp AC,$ $DM \perp BC,$ and $DN \perp AB.$ Prove that $DN^{2} = DM \cdot AN.$