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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 ΔPQR is double the perimeter of
ΔABC
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Solution
Clearly ABCQ and ARBC are parallelograms.
∴ BC = AQ and BC = AR
⇒ AQ = AR
⇒ A is the midpoint of QR .
Similarly B and C are the midpoints of PR and PQ respectively
∴ AB = `1/2` PQ, BC = `1/2` QR, CA = `1/2 `PR
⇒ PQ = 2AB,QR = 2BC and PR = 2CA
⇒ PQ + QR + RP = 2( AB + BC + CA)
⇒ Perimeter of DPQR = 2 [Perimeter of DABC ]
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