### 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.

**Answer:**

**Step by Step Explanation:**

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

Therefore, AE || FC ------(1)

AB = DC**[Opposite sides of the parallelogram ABCD.]**

AB =1 2

DC1 2 **[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.