Question
$ABC$ 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 $\triangle\text{PQR}$ is double the perimeter of $\triangle\text{ABC}.$
✓
Answer
Clearly $ABCQ$ and $ARBC$ are parallelograms.
Therefore, $BC = AQ$ and $BC = AR$
$\Rightarrow AQ = AR $
$\Rightarrow A$ is the mid-point of $QR$ Similarly $B$ and $C$ are the mid points of $PR$ and $PQ$ respectively.
$\therefore\text{AB}=\Big(\frac{1}{2}\Big)\text{PQ},$
$\text{BC}=\Big(\frac{1}{2}\Big)\text{QR},$
$\text{CA}=\Big(\frac{1}{2}\Big)\text{PR}.$
$\Rightarrow PQ = 2AB, QR = 2BC$ and $PR = 2CA $
$\Rightarrow PQ + QR + RP = 2 (AB + BC + CA)$
$\Rightarrow$ Perimeter of $\triangle\text{PQR}=2$
$($perimeter of $\triangle\text{ABC})$
Therefore, $BC = AQ$ and $BC = AR$
$\Rightarrow AQ = AR $
$\Rightarrow A$ is the mid-point of $QR$ Similarly $B$ and $C$ are the mid points of $PR$ and $PQ$ respectively.
$\therefore\text{AB}=\Big(\frac{1}{2}\Big)\text{PQ},$
$\text{BC}=\Big(\frac{1}{2}\Big)\text{QR},$
$\text{CA}=\Big(\frac{1}{2}\Big)\text{PR}.$
$\Rightarrow PQ = 2AB, QR = 2BC$ and $PR = 2CA $
$\Rightarrow PQ + QR + RP = 2 (AB + BC + CA)$
$\Rightarrow$ Perimeter of $\triangle\text{PQR}=2$
$($perimeter of $\triangle\text{ABC})$
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