Through $A, B$ and $C$,lines $RQ, PR$ and $QP$ have been drawn,respectively parallel to sides $BC, CA$ and $AB$ of a $\triangle ABC$ as shown in the figure. Show that $BC = \frac{1}{2} QR$.

(N/A) Given: $\triangle ABC$ and $\triangle PQR$ in which $AB \parallel PQ$,$BC \parallel RQ$ and $CA \parallel PR$.
To prove: $BC = \frac{1}{2} QR$
Proof: Consider the quadrilateral $ARBC$.
Since $AR \parallel BC$ (as $RQ \parallel BC$) and $RB \parallel AC$ (as $PR \parallel AC$),$ARBC$ is a parallelogram.
Therefore,$AR = BC$ (opposite sides of a parallelogram are equal) ... $(1)$
Now,consider the quadrilateral $ABCQ$.
Since $AQ \parallel BC$ (as $RQ \parallel BC$) and $QC \parallel AB$ (as $QP \parallel AB$),$ABCQ$ is a parallelogram.
Therefore,$AQ = BC$ (opposite sides of a parallelogram are equal) ... $(2)$
Adding equations $(1)$ and $(2)$,we get:
$AR + AQ = BC + BC$
$QR = 2BC$
$BC = \frac{1}{2} QR$
Hence,proved.