Question

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

Answer 1

Given in $\Delta ABC$, $PQ\parallel AB$ and $PR\parallel AC$ and $RQ\parallel BC$
To show $BC=\frac{1}{2}QR$.
Proof: In Quadrilateral $BCAR,BR\parallel CA$ and $BC\parallel RA$
So, quadrilateral , $BCAR$ is a parallelogram.
$\Rightarrow BC=AR$ ….(i)
Now, in quadrilateral $BCQA,BC\parallel AQ$ and $AB\parallel QC$
So, quadrilateral $BCQA$ is a parallelogram.
$\Rightarrow BC=AQ$ …..(ii)
On adding Eqs (i) and (ii) we get
$\Rightarrow 2BC=AR+AQ$
$\Rightarrow 2BC=RQ$
$\Rightarrow BC=\frac{1}{2}QR$.

Hence proved