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# ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see figure). Show that F is the mid-point of BC.

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.