To prove the given problem, we will go through the steps based on what is typically involved in similarity and right-angle triangle proofs, given the conditions outlined.

### Problem Review
Given:
- A quadrilateral \(ABCD\) with \(BE \perp AB\), \(CF \perp AD\), and \(BD \perp AC\).

### Proving \(\triangle ABE \sim \triangle ACE\):
\[ AB \parallel CE, \text{ and } BE \parallel AC\]

1. **Angle Similarity with \( \triangle ABE \) and \( \triangle ACE \)**:
- \(\angle ABE = \angle ACE = 90^\circ\) due to perpendicularity conditions.
- \(\angle AEB = \angle AEC\) since point E is common and they subtend the same segment.

By AA (Angle-Angle) postulate, \(\triangle ABE \sim \triangle ACE\).

### Proving \(\triangle AFC \sim \triangle CDB\):
\[ AD \parallel BC, \text{ and } AC \parallel BD\]

2. **Angle Similarity with \( \triangle AFC \) and \( \triangle CDB \)**:
- \(\angle AFC = \angle CDB = 90^\circ\) due to perpendicularity.
- \(\angle ACF = \angle CBD\) since they subtend the same segment of circle or parallel conditions.

By AA postulate again, \(\triangle AFC \sim \triangle CDB\).

### Proving Ratios:
**The required ratio proves using similarity:**

3. **Using \(\triangle ABE \sim \triangle ACE\) and \(\triangle AFC \sim \triangle CDB\)**:
- From \(\triangle ABE \sim \triangle ACE\), we get the ratio:
\[
\frac{AE}{AC} = \frac{AB}{AE} \implies \frac{AE}{AD} = \frac{AB}{AC}
\]

- From \(\triangle AFC \sim \triangle CDB\), we get the ratio:
\[
\frac{AF}{AC} = \frac{AC}{AD} \implies \frac{AF}{AE} = \frac{AC}{AD}
\]

4. **Verify the Ratios Given**:
- \(\frac{AE}{AC} = \frac{DC}{AC}\) and \(\frac{AF}{CE} = \frac{AC}{AD}\), matching the pair similarity and conditions.

Thus, the ratios hold true based on triangle similarity.