Question

$P$ is the midpoint of side $AB$ of a parallelogram $ABCD$. A line through $B$ parallel to $PD$ meets $DC$ at $Q$ and $AD$ produced at $R$. prove that (i) $AR=2BC$
(ii) $BR=2BQ$

Answer 1

$\left(i\right)$ In $△ARB,$ $P$ is the mid-point of $AB$ and $PD\parallel BR$.

$\therefore$ $D$ is a mid-point of $AR$ [ Converse mid point theorem ]

$\therefore$ $AR=2AD$

But $BC=AD$ [ Opposite sides of parallelogram are equal ]

$\therefore$ $AR=2BC$

$\left(ii\right)$ $ABCD$ is a parallelogram.

$\therefore$ $DC\parallel AB\Rightarrow DQ\parallel AB$

Now, in $△ARB,$

$D$ is a mid-point of $AR$ and $DQ\parallel AB$

$\therefore$ $Q$ is a mid point of $BR$ [ Converse mid-point theorem ]

$\Rightarrow$ $BR=2BQ$