ABCD is a parallelogram. E and F are the mid points of the sides AB and CD respectively. Prove that the segments CE and AF trisect the diagonal BD.
Asked by Topperlearning User | 16th Aug, 2017, 03:10: PM
AE||CF and AE = CF (halves of opposite sides of a parallelogram)
Therefore, AECF is a parallelogram.
In DPC, FQ||CP and F is the mid point of DC
So, PQ = QD
(Line through the mid point of one side of triangle and parallel to the another side, bisects the third side)
Similarly in ABQ, BP = PQ
Hence, BP = PQ = QD
Thus, CE and AF trisect the diagonal BD.
Answered by | 16th Aug, 2017, 05:10: PM
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