In $\Delta DEF, m \angle D = 90^{\circ}$. If $EF = 6$ and $DF = 4$,then $DE = \dots$

As shown in the adjoining diagram,$\angle B$ is an obtuse angle and $\overline{AM}$ is an altitude in $\Delta ABC$. Prove that $AC^{2} = AB^{2} + BC^{2} + 2 \cdot BC \cdot BM$.

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AM = 12$ and $CM = 3$,then $BM = \dots$

$\Delta ABC$ and $\Delta PQR$ are equilateral triangles. If $\frac{AB}{PQ} = \frac{3}{2}$,prove that $4 \times \text{Area of } \Delta ABC = 9 \times \text{Area of } \Delta PQR$.

In $\Delta XYZ$,$m\angle Y = 90^{\circ}$ and $\overline{YM}$ is an altitude to the hypotenuse $\overline{XZ}$. If $YM = 12$ and $XM = 8$,find $XZ$.

In $\Delta ABC$,$AB = AC$ and $m \angle A = 90^\circ$. If $BC = \sqrt{2} a$,then the area of $\Delta ABC$ is $\ldots \ldots \ldots \ldots$ $(a \in R^+)$.