In $\triangle ABC$ the median $AM$ is drawn. The foot of perpendicular from $B$ to the angle bisector of $\angle BMA$ is $B_1$ and the foot of perpendicular from $C$ to the angle bisector of $\angle AMC$ is $C_1.$ Let $MA$ and $B_1C_1$ intersect at $A_1.$ Find $\frac{B_1A_1}{A_1C_1}.$
1. **Define the problem and setup the geometry:**
   - Consider $\triangle ABC$ with median $AM$.
   - Let $B_1$ be the foot of the perpendicular from $B$ to the angle bisector of $\angle BMA$.
   - Let $C_1$ be the foot of the perpendicular from $C$ to the angle bisector of $\angle AMC$.
   - Let $MA$ and $B_1C_1$ intersect at $A_1$.
   - We need to find $\frac{B_1A_1}{A_1C_1}$.

2. **Introduce point $D$ on $AM$:**
   - Let $D$ be a point on $AM$ such that $MD = MC = MB$.
   - This implies $D$ is the midpoint of $BC$ since $M$ is the midpoint of $BC$ and $MD = MB = MC$.

3. **Reflection properties:**
   - Under reflection over $MC_1$, point $C$ maps to $D$ because $MC_1$ is the angle bisector and $C_1$ is the foot of the perpendicular from $C$.
   - Similarly, under reflection over $MB_1$, point $B$ maps to $D$.

4. **Rectangle formation:**
   - Since $D$ is the reflection of both $B$ and $C$ over $MB_1$ and $MC_1$ respectively, $D$ lies on both $BB_1$ and $CC_1$.
   - $\angle B_1MC_1 = \frac{\angle BMA + \angle AMC}{2} = 90^\circ$ because $B_1$ and $C_1$ are on the angle bisectors.
   - $\angle MB_1D = 90^\circ$ and $\angle MC_1D = 90^\circ$ imply that $MC_1DB_1$ is a rectangle.

5. **Intersection and ratio:**
   - Since $MC_1DB_1$ is a rectangle, $B_1C_1$ is parallel to $MD$ and $B_1C_1$ is bisected by $A_1$.
   - Therefore, $A_1$ is the midpoint of $B_1C_1$.

6. **Conclusion:**
   - Since $A_1$ is the midpoint of $B_1C_1$, the ratio $\frac{B_1A_1}{A_1C_1} = 1$.

The final answer is $\boxed{1}$.