Question

In the given below, $E$ is the mid-point of the side $AD$ of the trapezium $ABCD$ with $AB$ parallel to $DC$. A line through $E$ drawn to $AB$ intersects $BC$ at $F$. Show that $F$ is the mid-point of $BC$.

Answer 1

$ABCD$ is a trapezium in which $AB\parallel DC,$ $BD$ is a diagonal and $E$ is the mid-point of $AD$ and a line drawn through $E$ parallel to $AB$ intersecting $BC$ at $F$ such that $EF\parallel AB.$

$\Rightarrow$ Let $EF$ intersected $BD$ at $G.$

In $△ABD,$

$E$ is the mid-point of $AD$ and also $EG\parallel AB.$

We get, $G$ is the mid point of $BD$ [ By converse of mid point theorem ]

Similarly,

In $△BDC,$

$G$ is mid point of $BD$ and $GF\parallel AB\parallel DC.$

$\therefore$ $F$ is mid point of $BC$ [ By converse of mid point theorem ]