1. **State the problem:** Given that $HF \parallel BE$, $\angle HEF$ and $\angle BDC$ are complementary, and $\angle BDE$ and $\angle BDC$ are complementary, prove that $$\frac{HF}{BE} = \frac{HE}{BD}.$$\n\n2. **Given:**\n- $HF \parallel BE$\n- $\angle HEF$ and $\angle BDC$ are complementary\n- $\angle BDE$ and $\angle BDC$ are complementary\n\n3. **Understand complementary angles:**\nTwo angles are complementary if their measures add up to $90^\circ$.\n\n4. **From the given, since $\angle HEF$ and $\angle BDC$ are complementary, and $\angle BDE$ and $\angle BDC$ are complementary, it follows that:**\n$$\angle HEF = \angle BDE$$\nbecause both are complementary to $\angle BDC$.\n\n5. **Since $HF \parallel BE$, and $HE$ is a transversal, $\angle HEF$ and $\angle EBD$ are corresponding angles, so:**\n$$\angle HEF = \angle EBD.$$\n\n6. **From steps 4 and 5, we have:**\n$$\angle BDE = \angle EBD.$$\n\n7. **Triangles $HEF$ and $BDE$ share angles such that:**\n- $\angle HEF = \angle BDE$ (from step 4)\n- $\angle EHF = \angle BED$ (alternate interior angles because $HF \parallel BE$)\n- $\angle EFH = \angle BDE$ (from step 6)\n\n8. **By AA similarity criterion, triangles $HEF$ and $BDE$ are similar:**\n$$\triangle HEF \sim \triangle BDE.$$\n\n9. **From similarity, corresponding sides are proportional:**\n$$\frac{HF}{BE} = \frac{HE}{BD} = \frac{EF}{DE}.$$\n\n10. **Therefore, the required ratio is proven:**\n$$\frac{HF}{BE} = \frac{HE}{BD}.$$