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

In the figure,$ABC$ is a triangle in which $\angle ABC > 90^{\circ}$ and $AD \perp CB$ produced. Prove that $AC^{2} = AB^{2} + BC^{2} + 2BC \cdot BD$.

In $\Delta ABC$,$\overline{AD}$ and $\overline{BE}$ are medians. Line $m$ through $D$ and parallel to $\overline{BE}$ intersects $\overline{AC}$ at $K$. If $EK = 2.5$,then $AC = \ldots$

$ABCD$ is a trapezium in which $AB \parallel DC$ and $P$ and $Q$ are points on $AD$ and $BC$ respectively such that $PQ \parallel DC$. If $PD = 18 \, cm$,$BQ = 35 \, cm$ and $QC = 15 \, cm$,find $AD$ (in $cm$).

In $\Delta PQR$,$m\angle Q = 90^{\circ}$ and $\overline{QM}$ is an altitude to the hypotenuse $PR$. If $PQ = 8$ and $QR = 6$,find the area of $\Delta QMR$.

In $\Delta ABC$,$\overline{AD}$ is a median. The bisectors of $\angle ADB$ and $\angle ADC$ intersect $\overline{AB}$ and $\overline{AC}$ at $E$ and $F$ respectively. Prove that $\overline{EF} \parallel \overline{BC}$.