ABCD is a parallelogram in which E and F are the midpoints of the sides AB and CD respectively. Prove that the line segments CE and AF trisect the diagonal BD.

Explanation

Let BD be intersected by CE and AF at P and Q respectively.
AB||DC and `AB = DC" " `(opposite sides of a ||gm)
` rArr " AE||FC and " (1)/(2) AB =(1)/(2)DC rArr "AE|| FC and " AE = FC`
`rArr " AECF is a ||gm " rArr "AF||CE "rArr "EP ||AQ and FQ||CP." `
In `triangle ` BAQ, E is the midpoint of AB and EP||AQ, so P is the midpoint of BQ.
`therefore BP = PQ.`
Again, in `triangle `DPC, F is hte midpoint of DC and FQ||CP, so Q is the miodpoint of DP.
`therefore PQ = QD.`
`therefore BP= PQ=QD.`
Hence, CE and AF trisect the diagonal BD.