Question

in a parallelogram ABCD E and F are the midpoints of sides ab and CD respectively show that the line segment EF and 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.

Answer 2

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