Question

. In a parallelogram ABCD, E and Fare the midpoints of the
sides AB and DC respectively. Show that the line segments
AF and EC trisect the diagonal BD.​

Answer 1

Answer:

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

hope it helpz.

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