In trapezium ABCD,AB parallel CD and E is the | vedclass

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(A) $1$. Join $AC$. Let $AC$ intersect $EF$ at point $G$.$2$. In $	riangle ADC$,$E$ is the midpoint of $AD$ and $EG \parallel DC$ (since $EF \paralle

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(A) $1$. Join $AC$. Let $AC$ intersect $EF$ at point $G$.
$2$. In $\triangle ADC$,$E$ is the midpoint of $AD$ and $EG \parallel DC$ (since $EF \parallel AB$ and $AB \parallel CD$). By the Converse of the Midpoint Theorem,$G$ is the midpoint of $AC$. Thus,$EG = \frac{1}{2}CD$.
$3$. In $\triangle ABC$,$G$ is the midpoint of $AC$ and $GF \parallel AB$. By the Converse of the Midpoint Theorem,$F$ is the midpoint of $BC$. Thus,$GF = \frac{1}{2}AB$.
$4$. Adding the two segments: $EF = EG + GF = \frac{1}{2}CD + \frac{1}{2}AB = \frac{1}{2}(AB + CD)$.
$5$. Hence,$F$ is the midpoint of $BC$ and $EF = \frac{1}{2}(AB + CD)$.

In trapezium $ABCD$,$AB \parallel CD$ and $E$ is the midpoint of $AD$. $A$ line drawn through $E$ and parallel to $AB$ intersects $BC$ at $F$. Prove that $F$ is the midpoint of $BC$ and $EF = \frac{1}{2}(AB + CD)$.

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