# 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 Fig. 8.30). Show that F is the mid-point of BC.

**Solution:**

Let EF intersect DB at G as shown below.

By converse of mid-point theorem, we know that a line drawn through the mid-point of any side of a triangle and parallel to another side bisects the third side.

In trapezium ABCD,

EF || AB and E is the mid-point of AD.

Therefore, G is the mid-point of DB. [Converse of mid-point theorem]

As EF || AB and AB || CD,

∴ EF || CD (Two lines parallel to the same line are parallel to each other)

In ΔBCD, GF || CD and G is the mid-point of line BD.

Therefore, by using the converse of mid-point theorem, F is the mid-point of BC.

**Video Solution:**

## 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 Fig. 8.30). Show that F is the mid-point of BC

NCERT Maths Solutions Class 9 Chapter 8 Exercise 8.2 Question 4

**Summary:**

If ABCD is a trapezium in which AB || CD, BD is a diagonal and E is mid-point of AD, a line is drawn through E parallel to AB intersecting BC at F, then F is the mid-point of BC.

**☛ Related Questions:**

- ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig. 8.29). AC is a diagonal. Show that:(i) SR || AC and SR = 1/2AC(ii) PQ = SR(iii) 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.
- ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
- ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show thati) D is the mid-point of ACii) MD ⊥ ACiii) CM = MA = 1/2 AB

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