Question

In the adjoining figure, $\Delta ABC$ 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 $\Delta PQR$ is double the perimeter of $\Delta ABC$.

Answer 1

Soln.
Perimeter of $△ABC=AB+BC+CA$ ...(i)
Perimeter of $△PQR=PQ+QR+PR$ ...(ii)

Now $BC\parallel QAandCA\parallel QB$
i.e., BCQA is a parallelogram.
$\therefore BC=QA$ ...(iii)
Similarly, $BC\parallel ARandAB\parallel CR$
i.e., BCRA is a parallelogram.
$\therefore BC=AR$ ...(iv)
But, QR = QA + AR
From (iii) and (iv), we get:
⇒ QR = BC + BC
⇒ QR = 2BC ​
$\therefore BC=12\times QR$
Similarly, $CA=\frac{1}{2}PQandAB=\frac{1}{2}PR$
From (i) and (ii), we have:
Perimeter of $△ABC=\frac{1}{2}QR+\frac{1}{2}PQ+\frac{1}{2}PR$
=$\frac{1}{2}PR+QR+PQ$
i.e., Perimeter of $△ABC=\frac{1}{2}\left(Perimeterof△PQR\right)$
∴ Perimeter of (Perimeter of \triangle​ PQR) \) = 2 ⨯ (Perimeter of \triangle​ ABC ) \)