To prove that \(\Delta AHB = \Delta AHC\), we can use the criteria of congruence (specifically, the Side-Angle-Side - SAS criterion).

### Steps to prove \(\Delta AHB \cong \Delta AHC\):

1. **Common Side**:
- The side \(AH\) is common to both triangles \(\Delta AHB\) and \(\Delta AHC\).
- So, \(AH = AH\).

2. **Right Angles**:
- Since \(H\) is the foot of the perpendicular from \(A\) to line \(BC\), we have:
- \(\angle AHB = \angle AHC = 90^\circ\).

3. **Base Segments**:
- Since \(H\) is on line \(BC\), we can say that \(BH\) and \(CH\) are segments on line \(BC\), where \(H\) is between \(B\) and \(C\).
- The lengths of segments \(BH\) and \(CH\) do not need to be equal for the triangles to be congruent in this case as we are already using the right angle and the common side.

### Conclusion:
By SAS, we can conclude that:
\[
\Delta AHB \cong \Delta AHC
\]

Thus, we've proved that the triangles \(\Delta AHB\) and \(\Delta AHC\) are congruent.

Feel free to ask if you need help with parts b and c!