Question

Suppose $E$, $F$ are the midpoints respectively of the oblique sides $PS$, $QR$ of the trapezium $PQRS$. Prove that $EF$ is parallel to $SR$ and $EF=\frac{PQ+RS}{2}$.

Answer 1

Given: Trapezium $PQRS$, the mid-points $E$ of $PS$, $F$ of $QR$, and $EF$ is drawn.
To prove: $EF\parallel SR$ and
$EF=\frac{PQ+RS}{2}$.
Construction: Draw $PF$ and extend it to meet $SR$ produced at $T$.
Proof: Consider $△PQF$ and $△TRF$. Observe
$QF=FR$ (given $F$ is the midpoint of $QR$);
$\angle 1=\angle 2$ (vertically opposite angles);
$\angle 3=\angle 4$ (alternate angles by the transversal $PT$).
Hence $△PQF\cong △TRF$, by ASA postulate. It follows that
$PF=FT$ and $PQ=RT$.
Using mid-point theorem for the triangle $PST$, we conclude that $EF\parallel ST$ and $EF=\frac{ST}{2}$. But
$ST=SR+RT=SR+PQ$.

$\therefore EF=\frac{PQ+RS}{2}$.