Question

A b c make triangle pqr is triangle is parallel .prove that 2(ab+bc+ca)

Answer 1

Sol:  Given: ABC is a triangle and through vertices A, B and C. Lines are drawn parallel to BC, AC and AB respectively intersecting at D, E and F.RTP: Perimeter of ∆DEF = 2(Perimeter of ∆ABC) Proof: BC || AD, AC || BD ⇒ ACBD is a parallelogram. So, AD = BC --------------- (1) BC || AE, AB || EC ⇒ AECB is a parallelogram. So, AE = BC --------------- (2)From (1) and (2), AD = AE. ⇒ A is the mid point of DE Similarly, B is the midpoint of DF and C is the midpoint of EF. Also AB = 1/2 (EF)    [Midpoint theorem] BC = 1/2 (DE)           [Midpoint theorem] AC = 1/2 (DF)           [Midpoint theorem] AB + BC + AC = 1/2 (EF) + 1/2 (DE) + 1/2 (DF)  (AB + BC + AC) = 1/2 (EF + DE+ DF)  (EF + DE+ DF) = 2 (AB + BC + AC) Hence, perimeter of ∆DEF is double the perimeter of ∆ABC.