$E$ and $F$ are respectively the midpoints of non-parallel sides $A D, B C$ of trapezium $A B C D .$ Prove that $E F ∥ A B .$

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Solution

$A B C D$ is a trapezium in which $A B | | D C$ and $E, F$ are mid point of $A D, B C$ respectively. Join $C E$ and produce it to meet $B A$ produced at $G .$

In $△ E D C$ and $△ E A G,$

$E D = E A$ [ $E$ is the mid point of $A D .$ ]
$∠ C E D = ∠ G E C$ [Vertically opposite angles]
$∠ E C D = ∠ E G A$ [Alternating angles]

So, by $A S A$ criteria of similarity,
$△ E D C ≅ △ E A G$

$\Rightarrow C D = G A$ and $E C = E G$ which implies that $E$ is the midpoint of $C G$ as well.

In $△ C G B,$
$E$ is a mid point $C G$ and $F$ is a midpoint of $B C .$
$\Rightarrow \frac{C E}{E G} = \frac{C F}{F B} = 1$

So, by the converse of basic proportionality theorem $E F ∥ A B .$

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E and F are respectively the midpoints of non-parallel sides AD,BC of trapezium ABCD. Prove that EF∥AB.

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