### ABCD is a parallelogram with E and F as the mid-points of side AB and CD. Prove that the line segment AF and CE trisect the diagonal BD.

Step by Step Explanation:

1. It is given that in a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively.
2. We have to prove that line segment AF and EC trisect the diagonal BD.
3. Proof: AB || DC   [Opposite sides of the parallelogram ABCD.]
Therefore, AE || FC ------(1)
AB = DC   [Opposite sides of the parallelogram ABCD.]

 1 2
AB =
 1 2
DC   [Halves of equal are equal.]
AE = CF ------(2)
AECF is a parallelogram.
Therefore, EC || AF ------(3)   [Opposite sides of the parallelogram ABCD.]
In ΔDBC, F is the midpoint of DC and FP || CQ   [Because, EC || AF]
P is the midpoint of DQ.
So, DP = PQ ------(4)   [By converse of midpoint theorem.]
Similarly, In BAP, BQ = PQ ------(5)
On comparing eq. (3), (4) and (5), we get:
DP = PQ = BQ
Hence, line segment AF and EC trisect the diagonal BD.