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