# Ex.8.2 Q4 Quadrilaterals Solution - NCERT Maths Class 9

## Question

\(ABCD\) is a trapezium in which \(AB\) \( \parallel \) \(DC\), \(BD\) 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 the given figure). Show that \(F\) is the mid-point of \(BC\).

## Text Solution

**What is known?**

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 \(F.\)

**What is unknown?**

How we can show that F is the mid-point of \(BC.\)

**Reasoning:**

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.

**Steps:**

Let \(EF\) intersect \(DB\) at \(G\).

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 \(\Delta {ABD,}\)

\(EF\) \( \parallel \) \(AB\) and \(E\) is the mid-point of \(AD\).

Therefore, \(G\) will be the mid-point of \(DB\).

As \(EF\) \( \parallel \) \(AB\) and \(AB\) \( \parallel \) \(CD\),

\(\therefore\) \(EF\) \( \parallel \) \(CD\) (Two lines parallel to the same line are parallel to each other)

In \(\Delta {BCD,}\) \(GF\) \( \parallel \) \(CD\) and \(G\) is the mid-point of line \(BD\). Therefore, by using converse of mid-point theorem, \(F\) is the mid-point of \(BC\).