In rhombus $ABCD$,$\overline{AC} \cap \overline{BD} = \{O\}$. Then the area of $\Delta OAB = \ldots \ldots \ldots \ldots \times$ area of $\square ABCD$.

If $\Delta ABC \sim \Delta XYZ$ under the correspondence $ABC \leftrightarrow XZY$,then $BC^2 : YZ^2 = \ldots \ldots \ldots$ (Wait,the question asks for $ABC : XYZ = BC^2 : \ldots$ which refers to the ratio of areas). Given $\Delta ABC \sim \Delta XZY$,the ratio of their areas is equal to the ratio of the squares of their corresponding sides. Thus,$\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta XZY)} = \frac{BC^2}{ZY^2}$.

In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$. If $AB = 5.2$,$AC = 10.4$,and $BD = 3.8$,find $BC$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BE}$ is a median. If $AB = 3.6$ and $BC = 4.8$,find $BE$.

In $\Delta ABC,$ $\overline{AD}$ is a median. If $AD=7,$ $AC=7$ and $BD=4,$ find the perimeter of $\Delta ABC.$

In $\Delta ABC,$ the midpoints of the sides are $P, Q, R.$ In $\Delta PQR,$ the midpoints of the sides are $X, Y, Z.$ If the area of $\Delta XYZ$ is $20,$ find the area of $\Delta PQR$ and $\Delta ABC.$