$ABCD$ 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 $F$. Show that $F$ is the mid-point of $BC$.
(N/A) In trapezium $ABCD$,$AB \parallel DC$. $E$ is the mid-point of $AD$. $EF$ is drawn parallel to $AB$. We have to prove that $F$ is the mid-point of $BC$.
Join $BD$.
In $\Delta DAB$:
Since $E$ is the mid-point of $AD$ [Given] and $EG \parallel AB$ [Since $EF \parallel AB$],by the converse of the mid-point theorem,$G$ is the mid-point of $BD$.
In $\Delta BDC$:
Since $G$ is the mid-point of $BD$ [Proved] and $GF \parallel DC$ [Since $AB \parallel DC$ and $EF \parallel AB$,therefore $GF \parallel DC$],by the converse of the mid-point theorem,$F$ is the mid-point of $BC$.
Join $BD$.
In $\Delta DAB$:
Since $E$ is the mid-point of $AD$ [Given] and $EG \parallel AB$ [Since $EF \parallel AB$],by the converse of the mid-point theorem,$G$ is the mid-point of $BD$.
In $\Delta BDC$:
Since $G$ is the mid-point of $BD$ [Proved] and $GF \parallel DC$ [Since $AB \parallel DC$ and $EF \parallel AB$,therefore $GF \parallel DC$],by the converse of the mid-point theorem,$F$ is the mid-point of $BC$.
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