In the triangle $ABC$ , let $H$ be the foot of the altitude from $A $. Let $E$ be the intersection point of the angle bisector from $B$ with the side $AC$. Knowing that $\angle BEA=45^o$, determine $\angle EHC$.
1. **Draw the necessary lines and identify key points:**
   - Draw the altitude from \( A \) to \( BC \), and let \( H \) be the foot of this altitude.
   - Draw the angle bisector of \( \angle B \) and let it intersect \( AC \) at \( E \).
   - Draw a line perpendicular to \( AC \) through \( E \), and let this line intersect \( BC \) at \( I \).

2. **Identify and use given angles:**
   - Given that \( \angle BEA = 45^\circ \), we also have \( \angle BEI = 45^\circ \) because \( E \) lies on the angle bisector of \( \angle B \).

3. **Establish congruence of triangles:**
   - Since \( \angle AEB = \angle BEI = 45^\circ \) and \( \angle ABE = \angle EBI \) (as \( E \) is on the angle bisector), triangles \( \triangle AEB \) and \( \triangle IEB \) are congruent by the Angle-Angle (AA) criterion.

4. **Determine equal segments:**
   - From the congruence \( \triangle AEB \cong \triangle IEB \), it follows that \( AE = EI \).

5. **Calculate angles in the cyclic quadrilateral:**
   - Since \( AE = EI \), \( \triangle AEI \) is isosceles with \( \angle EAI = \angle AIE = 45^\circ \).
   - Since \( \angle AHI = 90^\circ \) (as \( H \) is the foot of the altitude from \( A \)), points \( A, E, I, H \) are concyclic (they lie on the same circle).

6. **Use properties of cyclic quadrilaterals:**
   - In a cyclic quadrilateral, opposite angles sum to \( 180^\circ \). Therefore, \( \angle EAI = \angle EHI \).
   - Given \( \angle EAI = 45^\circ \), it follows that \( \angle EHI = 45^\circ \).

7. **Determine the desired angle:**
   - Since \( \angle EHI = 45^\circ \) and \( \angle AHC = 90^\circ \), we have \( \angle EHC = \angle EHI = 45^\circ \).

The final answer is \( \boxed{45^\circ} \).