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