Question

in a parallelogram ABCD E and F are the midpoints of sides ab and CD respectively show that the line segments EF and ac trisect the diagonal BD​

Answer 1

Given ABCD is a parallelogram

Hence AB || CD ⇒ AE || FC

Also AB = CD (Opposite sides of parallelogram ABCD)

⇒ AE = FC (Since E and F are midpoints of AB and CD)

In quadrilateral AECF, one pair of opposite sides are equal and parallel.

∴ AECF is a parallelogram. ⇒ AF || EC (Since opposite sides of a parallelogram are parallel)

In ΔDPC, F is the midpoint of DC and FQ || CP

Hence Q is the midpoint of DQ by converse of midpoint theorem.

⇒ DQ = PQ → (1)

Similarly, in ΔAQB, E is the midpoint of AB and EP || AQ

Hence P is the midpoint of DQ by converse of midpoint theorem.

⇒ BP = PQ → (2)

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

BP = PQ = DQ

Hence, the line segments AF and EC trisect the diagonal BD of parallelogram ABCD.