In a parallelogram ABCD , E and F are the mid Points of sides AB and CD respectively. Show that the line segment AF and EC trisects the diagonal BD.
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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.
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.
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