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A line is drawn through E parallel to AB intersecting BC at F (see figure). Show that F is the mid-point of BC.

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.

- ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see figure). AC is a diagonal. Show thati) SR || AC and SR = (1/2) AC ii) PQ = SRiii) PQRS is a parallelogram.
- ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rectangle.
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- Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
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