Find the value of $x$ for which $DE \parallel AB$ in the given figure.

In $\square ABCD$,$A-Q-B$,$A-P-C$ and $A-R-D$. Given $\overline{PQ} \parallel \overline{BC}$ and $\overline{PR} \parallel \overline{DC}$. Prove that $\frac{AR}{AD} = \frac{AQ}{AB}$.

In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$. If $AB : AC = 3 : 4$ and $BD = 4.5$,then $BC = \ldots$

In $\Delta ABC$,$m \angle B = 90^{\circ}$. If $AB = 6$ and $AC = 10$,find the area of $\Delta ABC$.

In $\Delta ABC$,$AB = 17$,$BC = 15$,and $AC = 8$. Then,the length of the median on the longest side of the triangle is $\ldots$.

In $\Delta ABC$,$m \angle A = 90^\circ$. If $b = 12$ and $c = 35$,then $a = \ldots$