Question.
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 the given figure). Show that F is the mid-point of BC.
Solution:
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 $\triangle \mathrm{ABD}$,
$E F \| A B$ and $E$ is the mid-point of $A D$.
Therefore, G will be the mid-point of DB.
As EF $\| A B$ and $A B \| C D$
$\therefore \mathrm{EF} \| \mathrm{CD}$ (Two lines parallel to the same line are parallel to each other)
In $\triangle B C D, G F \| C D$ and $G$ is the mid-point of line $B D$. Therefore, by using converse of mid-point theorem, $F$ is the mid-point of $B C$.
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 $\triangle \mathrm{ABD}$,
$E F \| A B$ and $E$ is the mid-point of $A D$.
Therefore, G will be the mid-point of DB.
As EF $\| A B$ and $A B \| C D$
$\therefore \mathrm{EF} \| \mathrm{CD}$ (Two lines parallel to the same line are parallel to each other)
In $\triangle B C D, G F \| C D$ and $G$ is the mid-point of line $B D$. Therefore, by using converse of mid-point theorem, $F$ is the mid-point of $B C$.