In the adjoining figure, 
is a triangle and through 
lines are drawn, parallel respectively to 
and 
intersecting at 
and 
Prove that the perimeter of 
is double the perimeter of 
Here, Perimeter of ∆ABC = AB + BC + CA
And Perimeter of ∆PQR = PQ + QR + PR
Given that BC || QA and CA || QB which means BCQA is a parallelogram.
∴ BC = QA …(1)
Similarly, BC || AR and AB || CR, which means BCRA is a parallelogram.
∴ BC = AR …(2)
But, QR = QA + AR
From 1 and 2,
QR = BC + BC
∴ QR = 2BC
∴ BC = 
QR
Similarly, CA = 
PQ and AB = 
PR
Now,
Perimeter of ∆ABC = AB + BC + CA
= 
QR + 
PQ + 
PR
= 
(PR + QR + PQ)
This states that,
Perimeter of ∆ABC = 
(Perimeter of ∆PQR)
∴ Perimeter of ∆PQR = 2 ⨯ Perimeter of ∆ABC
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