Question

In a $△ABC$, lines are drawn through $A,B$ and $C$ parallel to the sides $BC,CA$ and $AB$ respectively forming $△PQR$. Prove that the perimeter of $△PQR$ is double the perimeter of $△ABC$.

Answer 1

$\because AQ\parallel CBandAC\parallel QB$

$\therefore AQBC$ is a parallelogram

$\therefore BC=AQ$ (Opposite side of a parallelogram)

$\because AR\parallel BCandAB\parallel RC$

$\therefore ARCB$ is a parallelogram

$\therefore BC=AR$ (Opposite side of a parallelogram)

Hence $A$ is the midpoint of $QR$

Similarly $B$ and $C$ are midpoints of $PQ$ and $PR$ respectively

$\therefore AB=\frac{1}{2}PRBC=\frac{1}{2}QRCA=\frac{1}{2}PQ$

$2AB=PR2BC=QR2CA=PQ$

$PR+QR+PQ=2(AB+BC+CA)$

Therefore,

Perimeter of $△$ PQR=2[Perimeter of $△$ABC]