Show that the line segment which joins the midpoints of the oblique sides of a trapezium is parallel to the parallel sides.
Let the trapezium be ABCD with E and F as the mid points of AD and BC, respectively.
Produce AD and BC to meet at P.
$\ln \triangle P A B, D C \| A B$
Applying Thales' theorem, we get:
$\frac{P D}{D A}=\frac{P C}{C B}$
Now, $E$ and $F$ are the midpoints of $A D$ and $B C$, respectively.
$\Rightarrow \frac{P D}{2 D E}=\frac{P C}{2 C F}$
$\Rightarrow \frac{P D}{D E}=\frac{P C}{C F}$
Applying the converse of Thales' theorem in $\Delta P E F$, we get that $D C \| E F$.
Hence, $E F \| A B$.
Thus. EF is parallel to both AB and DC.
This completes the proof.
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