The altitudes of the triangle ${ABC}$ meet in the point ${H}$. You know that ${AB = CH}$. Determine the value of the angle $\widehat{BCA}$. 

1. **Identify the given information and the goal:**
   - The altitudes of triangle \(ABC\) meet at point \(H\).
   - \(AB = CH\).
   - We need to determine the value of \(\angle BCA\).

2. **Analyze the given information:**
   - Since \(H\) is the orthocenter, the altitudes \(AD\), \(BE\), and \(CF\) intersect at \(H\).
   - Given \(AB = CH\), we need to use this information to find \(\angle BCA\).

3. **Use the properties of the orthocenter and the given equality:**
   - Since \(H\) is the orthocenter, \(\angle ABH = \angle ACH = 90^\circ - \angle BAC\).
   - Let \(E\) be the foot of the altitude from \(B\) to \(AC\).

4. **Use the given equality \(AB = CH\):**
   - Since \(AB = CH\), we can use the properties of the triangles formed by the altitudes.
   - Consider triangles \(ABE\) and \(HCE\):
     - \(\angle ABE = \angle HCE = 90^\circ\).
     - \(AB = CH\) (given).
     - \(\angle AEB = \angle HEC\) (vertical angles).

5. **Prove the triangles are congruent:**
   - By the Angle-Side-Angle (ASA) criterion, \(\triangle ABE \cong \triangle HCE\).
   - Therefore, \(AE = HE\).

6. **Determine the angle \(\angle HAC\):**
   - Since \(AE = HE\), \(\triangle AHE\) is isosceles with \(AE = HE\).
   - Thus, \(\angle HAE = \angle HEA\).

7. **Calculate the angles:**
   - Since \(\angle HAE + \angle HEA + \angle AHE = 180^\circ\) and \(\angle AHE = 90^\circ\), we have:
     \[
     \angle HAE + \angle HEA = 90^\circ
     \]
   - Since \(\angle HAE = \angle HEA\), we have:
     \[
     2\angle HAE = 90^\circ \implies \angle HAE = 45^\circ
     \]

8. **Relate \(\angle HAE\) to \(\angle ACB\):**
   - Since \(\angle HAE = 45^\circ\) and \(\angle HAE = 90^\circ - \angle ACB\), we have:
     \[
     45^\circ = 90^\circ - \angle ACB \implies \angle ACB = 45^\circ
     \]

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