Question

In the adjoining figure, $ABC$ is triangle. Through $A,$ $B$ and $C$ lines are drawn parallel to $BC,$ $CA$ and $AB$ respectively, which forms a $\Delta PQR.$ Show that $2\left(AB+BC+CA\right)=PQ+QR+RP.$

Answer 1

From the figure, it is given that, Through $A,B$ and $C$ lines are drawn parallel to $BC,CA$ and $AB$ respectively.
We have to show that $2(AB+BC+CA)=PQ+QR+RP$
$\because AB\parallel RC$ and $AR\parallel CB$
Therefore, $ABCR$ is a parallelogram.
$⟹AB=CR$ and $CB=AR$ (opposite sides of a parallelogram are equal) -------(1)
Similarly, $ABPC$ is a parallelogram $(\because AB\parallel CP$ and $PB\parallel CA)$

$⟹AB=PC$ and $AC=PB$ (opposite sides of a parallelogram are equal) -------(2)
Similarly, $ACBQ$ is a parallelogram $(\because BC\parallel AQ$ and $AC\parallel BQ)$

$⟹AC=BQ$ and $BC=AQ$ (opposite sides of a parallelogram are equal) --------(3)
By adding the equations (1), (2) and (3) we get,
$AB+AB+BC+BC+AC+AC=PB+PC+CR+AR+BQ+AQ$
$2AB+2BC+2AC=PQ+QR+RP$
By taking common we get,
$2(AB+BC+AC)=PQ+QR+RP$