Let $ABC$ be a triangle with $\angle A = 60^o$. The points $M,N,K$ lie on $BC,AC,AB$ respectively such that $BK = KM = MN = NC$. If $AN = 2AK$, find the values of $\angle B$ and $\angle C$.

by Mahdi Etesami Fard
1. **Identify the given information and draw the necessary segments:**
   - Given $\angle A = 60^\circ$ in $\triangle ABC$.
   - Points $M, N, K$ lie on $BC, AC, AB$ respectively such that $BK = KM = MN = NC$.
   - $AN = 2AK$.

2. **Analyze the segments and triangles formed:**
   - Join $KN$ and observe that $\triangle KMN$ is equilateral because $KM = MN$ and $\angle KMN = 60^\circ$.
   - Let $X$ be the midpoint of $AN$. Since $AN = 2AK$, it follows that $AX = AK = XN$.

3. **Examine $\triangle AKX$:**
   - Since $AK = AX = XN$, $\triangle AKX$ is equilateral.
   - Therefore, $\angle KAX = \angle AXK = \angle XKA = 60^\circ$.

4. **Analyze $\triangle KXN$:**
   - Since $KX = XN$ and $\angle KXN = 120^\circ$ (because $\angle KXA = 60^\circ$ and $\angle AXN = 60^\circ$), $\triangle KXN$ is isosceles.
   - Thus, $\angle XKN = \angle XNK = 30^\circ$.

5. **Determine $\angle XNM$:**
   - $\angle XNM = \angle XNK + \angle KNM = 30^\circ + 60^\circ = 90^\circ$.

6. **Analyze $\triangle NMC$:**
   - Since $\triangle NMC$ is isosceles with $MN = NC$ and $\angle XNM = 90^\circ$, it follows that $\angle MNC = 90^\circ$.
   - Therefore, $\angle ACM = 45^\circ$.

7. **Determine the angles in $\triangle ABC$:**
   - In $\triangle ABC$, $\angle A = 60^\circ$, $\angle C = 45^\circ$.
   - Using the angle sum property of a triangle, $\angle B = 180^\circ - \angle A - \angle C = 180^\circ - 60^\circ - 45^\circ = 75^\circ$.

Conclusion:
\[
\boxed{\angle B = 75^\circ \quad \text{and} \quad \angle C = 45^\circ}
\]