Given:
- M is the midpoint of side CD of parallelogram ABCD.
- Line BM intersects AC at L and AD produced at E.

To Prove:
EL = 2BL

Proof:

1. Drawing the figure:
- Draw a parallelogram ABCD.
- Mark the midpoint of side CD as M.
- Draw a line BM intersecting AC at L and AD produced at E.

2. Understanding the properties of a parallelogram:
- In a parallelogram, opposite sides are equal in length.
- In a parallelogram, opposite angles are equal.

3. Analyzing the given information:
- M is the midpoint of side CD, so CM = MD.
- BM intersects AC at L and AD produced at E.

4. Proving triangle BME and triangle CML are similar:
- In triangle BME and triangle CML, we have:
- BM shared by both triangles.
- ME and ML are parallel because they are corresponding sides of a parallelogram (ABCD).
- Angle EMB is equal to angle LMC as they are opposite angles in a parallelogram.
- Therefore, triangle BME and triangle CML are similar by the Angle-Angle (AA) criterion.

5. Using the similarity to find the ratio of corresponding sides:
- In similar triangles, the ratio of corresponding sides is equal.
- Therefore, we can write:
- (BL / ML) = (BM / ME)
- (BL / ML) = 1 / 2 (as M is the midpoint of CD)

6. Solving for BL:
- Cross-multiplying the above equation, we get:
- BL = ML / 2

7. Solving for EL:
- From the given information, we know that BM intersects AD produced at E.
- Therefore, angle AEM is equal to angle BML, as they are opposite angles formed by the transversal BM.
- Using the Angle-Angle (AA) criterion, we can conclude that triangle AEM and triangle BML are similar.
- In similar triangles, the ratio of corresponding sides is equal.
- Therefore, we can write:
- (EL / BL) = (EM / BM)
- Substituting the values, we get:
- (EL / BL) = 2 / 1 (as EM = 2BM)
- EL = 2BL

8. Conclusion:
- Hence, we have proved that EL = 2BL using the properties of a parallelogram and the similarity of triangles BME and CML.