$ \mathrm{ABCD} $ is a trapezium in which $ \mathrm{AB} \| \mathrm{DC}, \mathrm{BD} $ is a diagonal and $ \mathrm{E} $ is the mid-point of $ \mathrm{AD} $. A line is drawn through E parallel to $ \mathrm{AB} $ intersecting $ \mathrm{BC} $ at $ \mathrm{F} $ (see below figure). Show that $ \mathrm{F} $ is the mid-point of $ \mathrm{BC} $.
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Given:

\( \mathrm{ABCD} \) is a trapezium in which \( \mathrm{AB} \| \mathrm{DC}, \mathrm{BD} \) is a diagonal and \( \mathrm{E} \) is the mid-point of \( \mathrm{AD} \).

A line is drawn through E parallel to \( \mathrm{AB} \) intersecting \( \mathrm{BC} \) at \( \mathrm{F} \)

To do:
We have to show that \( \mathrm{F} \) is the mid-point of \( \mathrm{BC} \).

 Solution:

Let $P$ be the point of intersection of $BD$ and $EF$.

In $\triangle ABD$,

$EP \| AB$

$E$ is the mid-point of $AD$.

We know that,

A line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

Therefore,

$P$ is the mid-point of $BD$.

In $\triangle BCD$,

$PF \| CD$

$P$ is the mid-point of $BD$.

By converse of mid-point theorem,

$F$ is the mid-point of $CB$.

Hence proved.

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Updated on: 10-Oct-2022

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