In $\Delta PQR$,$m\angle Q = 90^{\circ}$ and $T$ is the midpoint of $\overline{PR}$. If $PQ = 6$ and $QR = 8$,then $QT = \ldots$

In $\Delta ABC$,the bisector of $\angle B$ intersects $\overline{AC}$ at $D$. If $\frac{AD}{DC} = \frac{3}{4}$ and $AB = 7.5$,then $BC = \ldots$

Which of the following correctly matches the information in Part $I$ and Part $II$?

Part $I$	Part $II$
$1.$ In $\Delta ABC$ and $\Delta PQR, \angle A \cong \angle P$ and $\angle C \cong \angle Q$	$a.$ Correspondence $ABC \leftrightarrow RQP$ is a similarity.
$2.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{QR} = \frac{BC}{PQ}$ and $\angle B \cong \angle Q$	$b.$ Correspondence $ABC \leftrightarrow QPR$ is a similarity.
$3.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{PQ} = \frac{BC}{PR} = \frac{CA}{QR}$	$c.$ Correspondence $ABC \leftrightarrow PQR$ is a similarity.
$4.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{PQ} = \frac{CA}{PR}$ and $\angle A \cong \angle P$	$d.$ Correspondence $ABC \leftrightarrow PRQ$ is a similarity.

Point $O$ lies in the interior of $\Delta PQR$. Given $O-A-P$,$O-B-Q$,$O-C-R$,$\overline{AB} \parallel \overline{PQ}$ and $\overline{BC} \parallel \overline{QR}$. Prove that $\overline{AC} \parallel \overline{PR}$.

In $\Delta PQR$,the bisector of $\angle P$ intersects $\overline{QR}$ at $S$. If $PQ = 8$,$QS = 5.6$ and $QR = 12.6$,find $PR$.

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