ABCD is a parallelogram with E and F as the mid-points of side AB and CD. Prove that the line segment AF and CE trisect the diagonal BD.


Answer:


Step by Step Explanation:

  1. It is given that in a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively.
  2. We have to prove that line segment AF and EC trisect the diagonal BD.
  3. Proof: AB || DC   [Opposite sides of the parallelogram ABCD.]
    Therefore, AE || FC ------(1)
    AB = DC   [Opposite sides of the parallelogram ABCD.]
     
    1
    2
     AB =  
    1
    2
     DC   [Halves of equal are equal.]
    AE = CF ------(2)
    AECF is a parallelogram.
    Therefore, EC || AF ------(3)   [Opposite sides of the parallelogram ABCD.]
    In ΔDBC, F is the midpoint of DC and FP || CQ   [Because, EC || AF]
    P is the midpoint of DQ.
    So, DP = PQ ------(4)   [By converse of midpoint theorem.]
    Similarly, In BAP, BQ = PQ ------(5)
    On comparing eq. (3), (4) and (5), we get:
    DP = PQ = BQ
    Hence, line segment AF and EC trisect the diagonal BD.

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