In $\Delta ABC$,$\overline{AD}$ is a median. Then,by Apollonius' theorem,$\ldots \ldots \ldots \ldots$ holds good.

$\square XYZW$ is a rectangle. If $XY + YZ = 17$ and $XZ + YW = 26$,find $XY$ and $YZ$ (given $XY > YZ$).

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 $\Delta ABC$,$D$ and $E$ are the midpoints of $\overline{BC}$ and $\overline{AC}$ respectively. $\overline{AD}$ and $\overline{BE}$ intersect at $G$. Line $m$ passing through $D$ and parallel to $\overline{BE}$ intersects $\overline{AC}$ at $K$. Then,$AC = \ldots$ (in $EK$)

In $\Delta XYZ$,$\overline{XN}$ is a median. If $XY^2 + XZ^2 = 200$ and $XN = 8$,find $YZ$.

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM : AB = 2 : 3$ and $AC = 15$,then $NC = \ldots$