In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is a median. If $AB = 20$ and $BC = 21$,find $BM$. (in $.5$)

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

Part $I$	Part $II$
$1.$ In $\Delta ABC$,$\angle B$ is a right angle and $\overline{BM}$ is a median.	$a. AB^2 + BC^2 = 2(BD^2 + CD^2)$
$2.$ In $\Delta ABC$,$\angle A$ is a right angle and $\overline{AD}$ is an altitude.	$b. BC = \frac{1}{2} AB$
$3.$ In $\Delta ABC$,$m\angle C = 90^\circ$ and $m\angle A = 30^\circ$.	$c. AC^2 = CD \cdot BC$
$4.$ In $\Delta ABC$,$\overline{BD}$ is a median.	$d. BM = \frac{1}{2} AC$

In the given figure,if $\angle 1 = \angle 2$ and $\triangle NSQ \cong \triangle MTR,$ then prove that $\triangle PTS \sim \triangle PRQ.$

In $\Delta ABC$,$\overline{AD}$ is a median and $AB^2 + AC^2 = 148$. If $AD = 7$,then $BC = \ldots$

In $\Delta ABC$,$m \angle B = 45^{\circ}$ and $\overline{AM}$ is an altitude,$M \in \overline{BC}$. If $BC = 7$ and $AM = 4$,then $AC = \ldots$

$\Delta ABC$ is an acute-angled triangle and $\overline{AM}$ is an altitude. Prove that $AC^{2} = AB^{2} + BC^{2} - 2 \cdot BC \cdot BM$.