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.
0.5/3

Answer 1

Answer:

BC=AQ ; BC=AR

Step-by-step explanation:

Clearly, ABCQ and ARBC are parallelograms.

∴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.

∴AB=

2

1

PQ,BC=

2

1

QR and CA=

2

1

PR

⇒PQ=2AB,QR=2BC and PR=2CA

⇒PQ+QR+RP=2(AB+BC+CA)

⇒ Perimeter of ΔPQR=2 (Perimeter of ΔABC)