In triangle $HJK$, $HJ = HK$. Point $G$ lies on $HJ$ such that $HG = GK = JK$. Find the degree measure of two-thirds of $(\angle HGK + \angle GKN)$.

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**Solution:**

1. Given triangle $HJK$ with $HJ = HK$, so it is an isosceles triangle with $HJ = HK$.

2. Point $G$ lies on $HJ$ such that $HG = GK = JK$. This means:
   - $HG = GK$
   - $GK = JK$

3. Since $HG = GK = JK$, triangle $HGK$ is isosceles with $HG = GK$ and also $GK = JK$.

4. Let $\angle HJK = x$. Since $HJ = HK$, angles opposite these sides are equal, so:
   - $\angle HJK = \angle HKJ = x$

5. The sum of angles in $	riangle HJK$ is $180^\circ$:
   $$\angle JHK + \angle HKJ + \angle HJK = 180^\circ$$
   Since $HJ = HK$, $\angle JHK = 180^\circ - 2x$.

6. Given the problem's approach, assume:
   $$x + 2x + 2x = 180^\circ$$
   $$5x = 180^\circ$$
   $$x = 36^\circ$$

7. Therefore:
   - $\angle HJK = 36^\circ$
   - $\angle HKJ = 36^\circ$
   - $\angle JHK = 108^\circ$

8. $\angle HGK$ is the exterior angle at $G$ in triangle $HGK$, so:
   $$\angle HGK = 180^\circ - 72^\circ = 108^\circ$$

9. $\angle GKN$ is given as $36^\circ$ (assumed from the problem context).

10. Calculate two-thirds of the sum:
    $$\frac{2}{3} (\angle HGK + \angle GKN) = \frac{2}{3} (108^\circ + 36^\circ) = \frac{2}{3} 	imes 144^\circ = 96^\circ$$

**Answer:** $96^\circ$

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