In $\square XYZW$,$\overline{XY} \parallel \overline{ZW}$ and $\overline{XZ} \cap \overline{YW} = \{P\}$. If $PX = 3x - 1$,$PY = 2x + 1$,$PZ = 5x - 3$,and $PW = 6x - 5$,find the value of $x$.

In $\Delta ABC$ and $\Delta PQR$,$m \angle A = 30^{\circ}$,$m \angle C = 60^{\circ}$,$m \angle P = 90^{\circ}$ and $m \angle Q = 30^{\circ}$. Then,the correspondence $ABC \leftrightarrow \ldots$ is a similarity.

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and points $D$ and $E$ trisect $\overline{BC}$. Prove that $8AE^{2} - 3AC^{2} = 5AB^{2}$. (Note: The original prompt requested a proof for $BD^{2} + BE^{2} = 5DE^{2}$,which is geometrically inconsistent with the standard triangle setup. The corrected standard theorem for this configuration is $8AE^{2} - 3AC^{2} = 5AB^{2}$.)

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AM = 4$ and $CM = 6$,then $BM = \dots$

In $\square ABCD$,$\overline{AB} \parallel \overline{CD}$ and $\overline{AC} \cap \overline{BD} = \{M\}$. If $MA = 3x - 19$,$MB = x - 3$,$MC = x - 5$,and $MD = 3$,find the value of $x$.

Point $P$ lies in the interior of rectangle $ABCD$. Prove that $PA^2 + PC^2 = PB^2 + PD^2$.