Answer: Yes
Step-by-step explanation:
Since ABCD is a rectangle, [tex]\angle AXY[/tex], [tex]\angle YBZ[/tex], [tex]\angle WCZ[/tex], and [tex]\angle WDX[/tex] are all right angles, and are thus all congruent because all right angles are congruent. Furthermore, because ABCD is a rectangle, we know that [tex]\overline{AB} \cong \overline{CD}[/tex] and [tex]\overline{AD} \cong \overline{BC}[/tex]. Because we are given that X, Y, Z, and W are midpoints, using the fact that halves of congruent segments are congruent, we can conclude that [tex]\overline{AY} \cong \overline{YB} \cong \overline{CW} \cong \overline{WD}[/tex] and that [tex]\overline{AX} \cong \overline{XD} \cong \overline{BZ} \cong \overline{ZC}[/tex]. As a result, we can conclude that [tex]\triangle AYX \cong \triangle DXW \cong \triangle CWZ \cong \triangle BYZ[/tex] by SAS, and thus by CPCTC, [tex]\overline{AY} \cong \overline{XW} \cong \overline{ZW} \cong \overline{YZ}[/tex]. Therefore, since the quadrilateral formed by the midpoints has four congruent sides, it must be a rhombus.
