L and M are the mid-points of sides AB and DC respectively of parallelogram ABCD. Prove that segments DL and BM trisect diagonal AC.

From figure,

As, ABCD is a parallelogram.

∴ AB = CD

⇒ $\dfrac{AB}{2} = \dfrac{CD}{2}$

⇒ BL = DM

As, AB || CD

∴ BL || DM

Since, in quadrilateral DLBM,

BL = DM and BL || DM

∴ DLBM is a parallelogram.

So we get, DL parallel to MB.

Suppose DL and BM intersect AC at P and Q respectively.

By converse of mid-point theorem,

The straight line drawn through the mid-point of one side of a triangle parallel to another, bisects the third side.

In triangle ABQ,

⇒ PL || QB (As, DL || MB)

⇒ Given, L is the midpoint of AB.

∴ P is the mid point of AQ (By converse of mid-point theorem)

∴ AP = PQ ……..(1)

In triangle CDP,

⇒ QM || PD (As, BM || DL)

⇒ Given, M is the midpoint of CD.

∴ Q is the mid point of CP (By converse of mid-point theorem)

∴ PQ = CQ ……..(2)

From equations (1) and (2), we get,

⇒ AP = PQ = CQ.

Hence, proved that DL and BM trisect the diagonal AC.