Question

In the figure, $\square ABCD$ is a trapezium, $AB\parallel DC$. Point $P$ and $Q$ are midpoints of $segAD$ and $segBC$ respectively. Then prove that, $PQ\parallel AB$ and $PQ=\frac{1}{2}(AB+DC)$.

Answer 1

Given,

$ABCD$ is a trapezium in which $AB\parallel DC$ and $P$, $Q$ are the midpoints of $AD$ & $BC$ respectively

Construction : Join $CP$ and produce it to meet $AB$ produced to $R$.

In $\Delta PDC$ & $\Delta PAR$

$PD=PA$ ($\because$ $P$ is the midpoint of $AD$)

$\angle CPD=\angle RPA$ (Vertically opposite angles)

$\angle PCD=\angle PRA$ (alternate angle)

$\therefore$ $\Delta PDC\cong \Delta PAR$ using $A$ as criterion

$\Rightarrow CP=CR$ $\left(CPCTC\right)$

Also,

In $\Delta CRB$,

$P$ is the midpoint of $CR$ (proved above)

Also, $Q$ is the midpoint of $BC$

$\Rightarrow$ By midpoint theorem $PQ\parallel AB$ and $PQ=\frac{1}{2}\left(RB\right)$

But $RB=RA+AB$

$=CD+AB$ ($\therefore$ $AR=CD$ by $CPCTC$)

$\therefore$ $PQ=\frac{1}{2}(CD+AB)$

Hence proved.