Question

In the adjoining figure, ∆ABC is a triangle and through A, B, C, lines are drawn, parallel respectively to BC, CA and AB, intersecting at P, Q and R. Prove that the perimeter of ∆PQR is double the perimeter of ∆ABC.

Answer 1

Perimeter of ∆​ABC = AB + BC + CA ...(i)
Perimeter of ∆PQR =​ PQ + QR + PR ...(ii)

BC || QA and CA || QB
i.e., BCQA is a parallelogram.
∴ BC = QA ...(iii)
Similarly, BC || AR and AB || CR
i.e., BCRA is a parallelogram.
∴ BC = AR ...(iv)
But, QR = QA + AR
From (iii) and (iv), we get:
⇒ QR = BC + BC
⇒ QR = 2BC​
∴ BC = $\frac{1}{2}QR$
Similarly, CA = ​$\frac{1}{2}PQ$ and AB = ​$\frac{1}{2}PR$

From (i) and (ii), we have:

Perimeter of ∆​ABC = $\frac{1}{2}QR$ + $\frac{1}{2}PQ$ + $\frac{1}{2}PR$

= $\frac{1}{2}\left(PR+QR+PQ\right)$

i.e., Perimeter of ∆​ABC = $\frac{1}{2}$ (Perimeter of ∆​PQR)
∴ Perimeter of ∆PQR = 2 ⨯ Perimeter of ∆ABC