To prove that triangles \( \triangle ABE \) and \( \triangle ACF \) are similar, we can break down the proof as follows:

### Step 1: Show that \( \triangle ABE \sim \triangle ACF \)

1. **Angle Correspondence**:
- Since \( AD \) and \( BE \) are altitudes, \( \angle ABE = 90^\circ \) and \( \angle ACF = 90^\circ \). Therefore, the angles at vertex \( A \) in both triangles are equal:
\[
\angle ABE = \angle ACF = 90^\circ
\]

2. **Using the properties of the triangle**:
- By the construction, we have:
- \( AD \) and \( HD \) being equal, and since they are both the heights from \( A \) to \( BC \), we have:
\[
AD = HD
\]

3. **Ratio of sides**:
- With \( BD \) forming part of \( AB \) and \( DC \) part of \( AC \), we have:
\[
\frac{AB}{AC} = \frac{BD}{DC}
\]

### Step 2: Ratios confirm similarity

Since both triangles have a right angle and one angle \( \angle ABE \) and \( \angle ACF \) are the same, by the Angle-Angle (AA) similarity criteria, we can conclude that:
\[
\triangle ABE \sim \triangle ACF
\]

### Conclusion

Thus, we have shown the similarity of the triangles \( \triangle ABE \) and \( \triangle ACF \) with respect to the angles and sides involved.