Given the conditions in the problem, we want to find the measure of angle $\angle XGH$ in the rectangle $EFGH$. Let's go through the solution step by step.

Step 1: Understanding the Geometry and Given Angles

1. Rectangle $EFGH$ is defined with vertices $E$, $F$, $G$, and $H$ in a counterclockwise order.
2. Point $M$ is the midpoint of side $EF$.
3. Point $X$ is such that $MH = MX$.
4. Given angles:
   • $\angle EFH = 32^\circ$.
   • $\angle MHX = 68^\circ$.

Step 2: Relate Given Angles to Geometry

Since $MH = MX$, triangle $MHX$ is isosceles with $\angle MHX = 68^\circ$. To find other angles in triangle $MHX$:

$\text{Let } \angle HMX = \angle MXH = x.$

Using the triangle angle sum property:

$\angle MHX + \angle HMX + \angle MXH = 180^\circ,$

$68^\circ + x + x = 180^\circ,$

$2x = 112^\circ,$

$x = 56^\circ.$

Thus, $\angle HMX = 56^\circ$.

Step 3: Finding $\angle XGH$

To find $\angle XGH$, we start by noting that $M$ lies on $EF$, and $M$ is the midpoint, so $\angle EMH = 90^\circ$.

Now, observe that:

$\angle EMX = \angle EMH - \angle HMX = 90^\circ - 56^\circ = 34^\circ.$

Since $\angle EFH = 32^\circ$, the angle $\angle XGH$ is given by:

$\angle XGH = \angle EMX + \angle EFH = 34^\circ + 32^\circ = 66^\circ.$

Final Answer

$\boxed{\angle XGH = 66^\circ}$

Would you like further explanation or have questions about this problem?

Related Questions

1. How can you verify that triangle $MHX$ is isosceles?
2. What is the significance of knowing that $M$ is the midpoint of $EF$?
3. Can $\angle XGH$ be found using an alternate geometric property?
4. How would the solution change if $\angle MHX$ were different?
5. What is the relationship between the angles within $\triangle MHX$ and the rest of the rectangle?

Tip

Remember that in problems involving rectangles, using properties like perpendicularity and midpoints can simplify complex angle calculations significantly!