Question

E and F are the mid point of AB and CD respectively of a parallelogram ABCD. After and CE intersect diagonal bd in P and Q respectively. Prove that diagonal BD trisect at P and Q

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.