In rectangle EFGH, let M be the midpoint of EF, and let X be a point such that MH=MX, as shown below. If if <EFH=32 and <MHX=68 then find <XGH in degrees.

Asymptote code below

[asy]
unitsize(1 cm);

pair A, B, C, D, M, X;

A = (0,3);
B = (4,3);
C = (4,0);
D = (0,0);
M = (A + B)/2;
X = rotate(26,M)*(D);

draw(A--B--C--D--cycle);
draw(D--M--X--cycle);
draw(C--X);
draw(B--D);

label("$E$", A, NW);
label("$F$", B, NE);
label("$G$", C, SE);
label("$H$", D, SW);
label("$M$", M, N);
label("$X$", X, S);
[/asy]

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!

In rectangle EFGH, let M be the midpoint of EF, and let X be a point such that MH=MX. If ∠EFH=32° and ∠MHX=68°, then find ∠XGH in degrees.

Learn how to solve for angle ∠XGH in a rectangle EFGH using properties of isosceles triangles and triangle angle sum theorems. Given angles ∠EFH=32° and ∠MHX=68°, this problem involves geometric principles suitable for students in Grades 9-11.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation