Question 1169822
Absolutely! Let's break down this geometry problem step-by-step.

**Understanding the Problem**

We're given a triangle DEF with specific conditions regarding the extensions of its sides and the lengths of certain segments. We need to prove that the angle EKG is equal to 90 - x degrees.

**Diagram**

It's crucial to visualize this problem. Here's a description of the diagram:

1.  Start with triangle DEF.
2.  Extend DE to G and DF to H.
3.  EF = EG = FH.
4.  EH and FG intersect at K.

**Solution**

1.  **Isosceles Triangles:**
    * Since EG = EF, triangle EFG is isosceles. Therefore, ∠EFG = ∠EGF.
    * Since FH = EF, triangle EFH is isosceles. Therefore, ∠FEH = ∠FHE.

2.  **Angle Sum of Triangle DEF:**
    * In triangle DEF, ∠DEF + ∠DFE + ∠EDF = 180°.
    * Let ∠DEF = a and ∠DFE = b.
    * Then, a + b + 2x = 180°.
    * Therefore, a + b = 180° - 2x.

3.  **Angles in Isosceles Triangles:**
    * In triangle EFG, let ∠EFG = ∠EGF = y.
    * In triangle EFH, let ∠FEH = ∠FHE = z.
    * The angles on a straight line add to 180 degrees.
    * ∠DEG = 180 - a, and ∠DFH = 180 -b.
    * In triangle EFG, 2y + 180-a = 180, so 2y = a, and y = a/2.
    * In triangle EFH, 2z + 180-b = 180, so 2z = b, and z = b/2.

4.  **Angles in Triangle EKF**
    * ∠EKF = 180 - (z + y)
    * ∠EKF = 180 - (b/2 + a/2)
    * ∠EKF = 180 - (a+b)/2

5.  **Substitution**
    * We know a + b = 180 - 2x
    * Substitute this into the equation for ∠EKF:
    * ∠EKF = 180 - (180 - 2x)/2
    * ∠EKF = 180 - (90 - x)
    * ∠EKF = 180 - 90 + x
    * ∠EKF = 90 + x

6.  **Vertical Angles**
    * ∠EKG and ∠EKF are supplementary angles, so ∠EKG + ∠EKF = 180.
    * ∠EKG = 180 - ∠EKF
    * ∠EKG = 180 - (90 + x)
    * ∠EKG = 180 - 90 - x
    * ∠EKG = 90 - x.

**Therefore, we have shown that angle EKG = 90 - x degrees.**