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Answer :
Given: ABCD is a trapezium in which AB || CD and E is mid-point of AD and EF || AB.
In \(\triangle{ABD}\), we have,
EP || AB and E is mid-point of AD.
So, by theorem, if a line drawn through the
mid-point of one side of a triangle is parallel to another side, then it bisects the third side.
Therefore, P is the midpoint of BD.
Similarly, in \(\triangle{BCD}\),
we have, PF || CD,
Therefore by converse of mid point theorem, F is mid-point of BD.