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.

Answer 1

ABCQ and ARBC are parallelograms. 

Therefore, BC = AQ and BC = AR 

⇒ AQ = AR 

⇒ A is the mid-point of QR 

Similarly B and C are the mid points of PR and PQ respectively. 

By mid−point theorem, we have 

AB = PQ/2, BC = QR/2 and CA = PR/2 

or PQ = 2AB, QR = 2BC and PR = 2CA 

⇒ PQ + QR + RP = 2 (AB + BC + CA) 

⇒ Perimeter of ΔPQR = 2 (Perimeter of ΔABC) 

Hence proved.