Through the midpoint of the side CD of parallelogram ABCD, the line BM is drawn intersecting AC at L and AD produced at E. We need to prove that EL is equal to 2BL.

Proof:

1. Given:
- ABCD is a parallelogram.
- BM is drawn through the midpoint of CD.
- BM intersects AC at L and AD produced at E.

2. To Prove:
EL = 2BL

3. Proof:

3.1. Construction:
- Draw a line segment BL parallel to AD.
- Let the point where BL intersects CD be F.

3.2. Explanation:
- Since ABCD is a parallelogram, opposite sides are parallel and equal in length.
- Therefore, AD || BC and AB || DC.
- Also, since BM is drawn through the midpoint of CD, it divides CD into two equal parts, CF and FD.
- Hence, CF = FD.

3.3. Proof of EL = 2BL:
- In triangle BCF and triangle BDF,
- BC = AD (opposite sides of parallelogram ABCD are equal)
- CF = FD (BM is drawn through the midpoint of CD)
- Angle BCF = Angle BDF (vertically opposite angles)
- Therefore, triangle BCF is congruent to triangle BDF (by SAS congruence)
- By the congruence of triangles, we can conclude that BF = BF.
- Since BF is a transversal, we have BL || AD.
- Therefore, triangle BEL and triangle BAD are similar (by AAA similarity).
- From the similarity of triangles, we can conclude that:
- EL/BL = AD/BD
- EL/BL = 1/2 (since AD = BD in a parallelogram)
- EL = 2BL

4. Conclusion:
- We have proved that EL is equal to 2BL using the given information and properties of parallelograms.

Hence, the statement is true.

5. Additional Information:
- Parallelograms have several properties, including opposite sides being parallel and congruent, opposite angles being equal, diagonals bisecting each other, and the sum of adjacent angles being 180 degrees.
- The midpoint of a line segment is a point that divides the line segment into two equal parts.
- In a triangle, SAS (side-angle-side) congruence and AAA (angle-angle-angle) similarity are two important criteria for determining congruence and similarity.