1. **Problem statement:** Given the proportion $\frac{AC}{AE} = \frac{AB}{AD}$, prove that $ED \parallel CB$. 2. **Step 1: Identify the common angle.** We have $\angle 8 = \angle \hat{A}$, which is the common angle in triangles $AED$ and $ABC$. 3. **Step 2: Prove similarity of triangles.** From the given proportion and the common angle, triangles $AED$ and $ABC$ are similar by the Side-Angle-Side (SAS) similarity criterion. 4. **Step 3: Corresponding angles in similar triangles.** Since $\triangle AED \sim \triangle ABC$, corresponding angles are equal, so $\angle E_1 = \angle C_2$. 5. **Step 4: Use the parallel line criterion.** If a transversal intersects two lines and the corresponding angles are equal, then the two lines are parallel. 6. **Conclusion:** Since $\angle E_1 = \angle C_2$, it follows that $ED \parallel CB$. This completes the proof.