Let $AH$ be the altitude of a given triangle $ABC.$ The points $I_b$ and $I_c$ are the incenters of the triangles $ABH$ and $ACH$ respectively. $BC$ touches the incircle of the triangle $ABC$ at a point $L.$ Find $\angle LI_bI_c.$
1. **Identify the given elements and their relationships:**
   - \( AH \) is the altitude of triangle \( \triangle ABC \).
   - \( I_b \) and \( I_c \) are the incenters of triangles \( \triangle ABH \) and \( \triangle ACH \), respectively.
   - \( BC \) touches the incircle of \( \triangle ABC \) at point \( L \).

2. **Define the feet of perpendiculars:**
   - Let \( D \) and \( E \) be the feet of the perpendiculars from \( I_b \) and \( I_c \) onto \( BC \), respectively.

3. **Establish the distances from the incenters to the sides:**
   - Since \( I_b \) is the incenter of \( \triangle ABH \), the distance from \( I_b \) to \( BC \) (denoted as \( DI_b \)) is equal to the distance from \( H \) to \( BC \) (denoted as \( DH \)).
   - Similarly, since \( I_c \) is the incenter of \( \triangle ACH \), the distance from \( I_c \) to \( BC \) (denoted as \( EI_c \)) is equal to the distance from \( H \) to \( BC \) (denoted as \( EH \)).

4. **Calculate the distances:**
   - The distance \( DH \) can be expressed as:
     \[
     2DH = AH + BH - AB
     \]
   - The distance \( EH \) can be expressed similarly:
     \[
     2EH = AH + CH - AC
     \]

5. **Relate the distances to the point \( L \):**
   - Since \( L \) is the point where \( BC \) touches the incircle of \( \triangle ABC \), we have:
     \[
     2LE = AC + CB - AB - AC - CH + AH = AH + BH - AB = 2DH
     \]
   - This implies:
     \[
     DL = HE = EI_c
     \]

6. **Establish the congruence of triangles:**
   - From the above relationships, we have:
     \[
     \triangle I_bDL \equiv \triangle LEI_c
     \]
   - This congruence implies that:
     \[
     \angle I_bLI_c = 90^\circ
     \]

The final answer is \( \boxed{ \angle I_bLI_c = 90^\circ } \)