Let $A_1$ be a midmoint of $BC$, and $G$ is a centroid of the non-isosceles triangle $\triangle ABC$. $GBKL$ and $GCMN$ are the squares lying on the left with respect to rays $GB$ and $GC$ respectively. Let $A_2$ be a midpoint of a segment connecting the centers of the squares $GBKL$ and $GCMN$. Circumcircle of triangle $\triangle A_{1}A_{2}G$ intersects $BC$ at points $A_1$ and $X$. Find $\frac{A_{1}X}{XH}$, where $H$ is a base of altitude $AH$ of the triangle $\triangle ABC$.
1. **Identify Key Points and Properties:**
   - Let \( A_1 \) be the midpoint of \( BC \).
   - \( G \) is the centroid of \( \triangle ABC \).
   - \( GBKL \) and \( GCMN \) are squares constructed on the left of rays \( GB \) and \( GC \) respectively.
   - Let \( F \) and \( E \) be the centers of squares \( GBKL \) and \( GCMN \) respectively.
   - \( A_2 \) is the midpoint of the segment connecting \( F \) and \( E \).

2. **Analyze the Geometry of the Squares:**
   - Since \( GBKL \) and \( GCMN \) are squares, the centers \( F \) and \( E \) are located at the midpoints of the diagonals of these squares.
   - The diagonals of a square are perpendicular bisectors of each other and intersect at the center of the square.

3. **Similarity of Triangles:**
   - Note that \( \triangle CGB \sim \triangle FGE \) with a similarity ratio of \( \sqrt{2} \). This is because the sides of the squares are \( \sqrt{2} \) times the length of the segments \( GB \) and \( GC \).

4. **Angle Calculation:**
   - Since \( \triangle CGB \sim \triangle FGE \), the angles are preserved. Therefore, \( \angle CGB = \angle FGE \).
   - Given that \( \angle CGB = 90^\circ \) (since \( GBKL \) and \( GCMN \) are squares), it follows that \( \angle FGE = 90^\circ \).

5. **Isosceles Right Triangle:**
   - Since \( \triangle CGB \sim \triangle FGE \) and the similarity ratio is \( \sqrt{2} \), \( \triangle GA_1A_2 \) is an isosceles right triangle with \( \angle A_1GA_2 = 45^\circ \).
   - Therefore, \( GA_1 = \sqrt{2} \cdot GA_2 \).

6. **Circumcircle and Intersection:**
   - Since \( \triangle GA_1A_2 \) is an isosceles right triangle, \( A_1G \) is the diameter of the circumcircle of \( \triangle A_1A_2G \).
   - This implies that \( \angle A_1XG = 90^\circ \) where \( X \) is the intersection of the circumcircle with \( BC \).

7. **Parallel Lines and Ratio Calculation:**
   - Since \( \angle A_1XG = 90^\circ \) and \( XG \parallel AH \), the triangles \( \triangle A_1XG \) and \( \triangle AHG \) are similar.
   - Therefore, the ratio \( \frac{A_1X}{XH} = \frac{A_1G}{GA} \).

8. **Centroid Properties:**
   - The centroid \( G \) divides the median \( AA_1 \) in the ratio \( 2:1 \).
   - Hence, \( \frac{A_1G}{GA} = \frac{1}{2} \).

The final answer is \( \boxed{ \frac{A_1X}{XH} = \frac{1}{2} } \).