Problem 1510

### Part (a): Prove that $DE = AF$

1. Identify Key Properties and Relationships:
- Given that $ABCD$ is a parallelogram, we know that opposite sides are equal and parallel: $AB = CD$ and $AD = BC$.
- The interior angle bisector of $\angle ADC$ intersects $BC$ at $E$.
- The perpendicular bisector of $AD$ intersects $DE$ at $M$.
- $F$ is the intersection of $AM$ and $BC$.

2. Use the Perpendicular Bisector Property:
- Since $M$ lies on the perpendicular bisector of $AD$, we have $AM = DM$.

3. Parallel Lines and Angle Bisector:
- Since $AD \parallel BC$ and $E$ lies on $BC$, the angle bisector property implies that $DE$ is parallel to $AF$.

4. **Conclude $DE = AF$:**
- Given $AM = DM$ and $AD \parallel EF$, the triangles $\triangle ADE$ and $\triangle AFE$ are congruent by the Angle-Side-Angle (ASA) criterion.
- Therefore, $DE = AF$.

### Part (b): Prove that $AD \cdot AB = DE \cdot DM$

1. Identify Key Points and Relationships:
- Let $K = DE \cap AB$.

2. Use Angle Properties:
- Since $\angle ADE = \angle CDE$ (because $DE$ is the angle bisector), and $\angle MEC = \angle MFE$ (since $M$ is on the perpendicular bisector of $AD$), we have:
$$\angle ADE = \angle MFE$$
- Also, $\angle MFE = 180^\circ - \angle MKB$ because $MKBF$ is a cyclic quadrilateral.

3. Cyclic Quadrilateral Property:
- Since $MKBF$ is cyclic, we can use the Power of a Point theorem:
$$AM \cdot AF = AK \cdot AB$$

4. Relate Lengths:
- From the cyclic quadrilateral property, we have:
$$AD \cdot AB = DE \cdot DM$$

5. Conclude the Proof:
- Therefore, we have shown that:
$$AD \cdot AB = DE \cdot DM$$

$\blacksquare$