Question

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

Answer 1

DM=LB (both are half of equal sides AB and CD)

and

DM paralle to LB (as AB and CD are parallel)

So, DLBM is a parallelogram.(one pair of opposite sides is parallel and equal)

So we get, DL parallel to MB.

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

Consider triangle ABC,

PL parallel to QB ( as DL parallel to MB)

L is the midpoint of AB,

so P must be the mid point of AQ( conv of mid pt thm)

So,

AP=PQ..(i)

Simly, considering triangle CDP,

we get

CQ=QP..(ii)

from (i) and (ii), we get,

AP=PQ=CQ.

Thus,

P and Q trisect the diagonal AC.
Like if satisfied