$ABC$ is a triangle and through $A, B, C$ lines are drawn parallel to $BC, CA$ and $AB$ respectively intersecting at $P, Q$ and $R$. Prove that the perimeter of $\triangle PQR$ is double the perimeter of $\triangle ABC$.
Given:
$ABC$ is a triangle and through $A, B, C$ lines are drawn parallel to $BC, CA$ and $AB$ respectively intersecting at $P, Q$ and $R$.
To do:
We have to prove that the perimeter of $\triangle PQR$ is double the perimeter of $\triangle ABC$.
Solution:
$PQ \parallel BC$ and $QR \parallel AB$
This implies,
$ABCQ$ is a parallelogram.
$BC = AQ$
Similarly,
$BCAP$ is a parallelogram.
$BC = AP$.....…(i)
$AQ = AP = BL$
This implies,
$PQ = 2BC$
Similarly,
$QR = 2AB$ and $PR = 2AC$
Therefore,
Perimeter of $\triangle PQR= PQ + QR + PR$
$= 2AB + 2BC + 2AC$
$= 2(AB + BC + AC)$
$= 2 \times$ perimeter of $\triangle ABC$
Hence proved.
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