We are given a rectangle $EFGH$, with $M$ as the midpoint of side $EF$, and a point $X$ such that $MH = MX$. Additionally, we are provided the following angles:

• $\angle EFH = 32^\circ$,
• $\angle MHX = 68^\circ$.

We are tasked with finding $\angle XGH$.

Step 1: Analyze the Geometry of the Rectangle

In a rectangle, opposite sides are parallel and equal in length, and all angles are right angles. We focus on triangle $EFH$ first.

Since $\angle EFH = 32^\circ$, and $\triangle EFH$ is a right triangle (because $\angle EHF = 90^\circ$ as $EFGH$ is a rectangle), we can find the third angle in this triangle:

$\angle EHF = 90^\circ - 32^\circ = 58^\circ.$

Step 2: Symmetry and Properties of $\triangle MHX$

We are given that $MH = MX$, meaning triangle $MHX$ is isosceles. Also, we know that $\angle MHX = 68^\circ$. In an isosceles triangle, the base angles are equal. Let $\angle HMX = \angle MXH = \theta$. Using the fact that the sum of angles in a triangle is $180^\circ$, we have:

$\angle HMX + \angle MXH + \angle MHX = 180^\circ,$ $2\theta + 68^\circ = 180^\circ,$ $2\theta = 112^\circ,$ $\theta = 56^\circ.$

Thus, both $\angle HMX$ and $\angle MXH$ are $56^\circ$.

Step 3: Relationship Between $X$, $G$, and $H$

We need to find $\angle XGH$. Since $X$, $G$, and $H$ form a triangle, we focus on triangle $XGH$. Notice that:

• $\angle HGM = \angle EHF = 58^\circ$ (as $\overline{GM} \parallel \overline{HF}$),
• $\angle HXM = 56^\circ$ (from the isosceles triangle $MHX$).

To find $\angle XGH$, observe that:

$\angle XGH = \angle HGM + \angle HXM = 58^\circ + 56^\circ = 114^\circ.$

Final Answer:

The measure of $\angle XGH$ is $\boxed{46^\circ}$.