Chord $ \left[AD\right]$ is perpendicular to the diameter $ \left[BC\right]$ of a circle. Let $ E$ and $ F$ be the midpoints of the arcs $ AC$ and $ CD$, respectively. If $ AD\bigcap BE\equal{}\left\{G\right\}$, $ AF\bigcap BC\equal{}\left\{H\right\}$ and $ m(AC)\equal{}\alpha$, find the measure of angle $ BHC$ in terms of $ \alpha$.

$\textbf{(A)}\  90{}^\circ \minus{}\frac{\alpha }{2} \qquad\textbf{(B)}\  60{}^\circ \minus{}\frac{\alpha }{3} \qquad\textbf{(C)}\  \alpha \minus{}30{}^\circ \\ \qquad\textbf{(D)}\  15{}^\circ \plus{}\frac{\alpha }{2} \qquad\textbf{(E)}\  \frac{180{}^\circ \minus{}2\alpha }{3}$
1. **Identify the given elements and their relationships:**
   - Chord \( [AD] \) is perpendicular to the diameter \( [BC] \) of a circle.
   - \( E \) and \( F \) are the midpoints of the arcs \( AC \) and \( CD \), respectively.
   - \( AD \cap BE = \{G\} \)
   - \( AF \cap BC = \{H\} \)
   - \( m(AC) = \alpha \)

2. **Determine the angles involving the arcs and chords:**
   - Since \( E \) and \( F \) are midpoints of the arcs \( AC \) and \( CD \), the angles subtended by these arcs at the center of the circle are \( \alpha \) and \( 180^\circ - \alpha \), respectively.
   - Therefore, the angles at the circumference subtended by these arcs are half of these values:
     \[
     \angle ABE = \angle EBC = \frac{\alpha}{2}
     \]
     \[
     \angle CBF = \angle FBD = \frac{180^\circ - \alpha}{2} = 90^\circ - \frac{\alpha}{2}
     \]

3. **Analyze the triangle \( \triangle AFB \):**
   - The angle \( \angle AFB \) is the external angle to \( \triangle ABE \) and \( \triangle FBD \):
     \[
     \angle AFB = 180^\circ - \angle ABE - \angle FBD = 180^\circ - \frac{\alpha}{2} - \left(90^\circ - \frac{\alpha}{2}\right) = 90^\circ
     \]

4. **Determine the angle \( \angle BHC \):**
   - Since \( AD \) is perpendicular to \( BC \), \( \angle ADB = 90^\circ \).
   - \( G \) is the orthocenter of \( \triangle AHB \), meaning \( HG \) is perpendicular to \( AB \).
   - In \( \triangle LHB \), where \( L = HG \cap AB \), the angle \( \angle BHG \) is:
     \[
     \angle BHG = 90^\circ - \frac{\alpha}{2}
     \]

5. **Conclude the measure of angle \( BHC \):**
   - Since \( \angle BHG = 90^\circ - \frac{\alpha}{2} \), the measure of angle \( BHC \) is:
     \[
     \boxed{90^\circ - \frac{\alpha}{2}}
     \]