Question

In a ΔABC, AD, BE and CF are the medians drawn from the vertices A, B and C respectively. Then study the following statements and choose the correct option.

I. 3 (AB + BC + AC) > 2 (AD + BE + CF) 

II. 3 (AB + BC + AC) < 2 (AD + BE + CF) 

III. 3 (AB + BC + AC) < 4 (AD + BE + CF) 

IV. 3 (AB + BC + AC) > 4 (AD + BE + CF) 

(a) I and IV are true 

(b) I and III are true 

(c) II and IV are true 

(d) II and III are true

Answer 1

(b) I and III are true

G being the centroid of ΔABC,

\(\frac{AG}{GD}=\frac{BG}{GE}=\frac{CG}{GF}=\frac{2}{1}\)

In ΔABD, 

AB + BD > AD

⇒ AB + \(\frac{BC}{2}\) > AD           ....(i)

In ΔBEC, 

BC + CE > BE ⇒ BC + \(\frac{AC}{2}\) > BE        ....(ii)

In ΔAFC, 

AC + AF > CF ⇒ AC + \(\frac{AB}{2}\) > CF        .....(iii)

Adding (i), (ii) and (iii), we get

AB + BC + AC + \(\frac{BC}{2}\) + \(\frac{AC}{2}\) + \(\frac{AB}{2}\) > AD + BE + CF

⇒ \(\frac{2AB+2BC+2AC+BC+AC+AB}{2}\) > AD + BE + CF

⇒ 3(AB + BC + AC) > 2 (AD + BE + CF) ⇒ I is true Also, in Δ BGC, 

BG + GC > BC

\(\big(∵ \frac{BG}{GE}=\frac{2}{1}\) and \(\frac{CG}{GF}=\frac{2}{1}\) ⇒ BG = \(\frac{2}{3}\)BE and CG = \(\frac{2}{3}\)CF\(\big)\)

⇒ \(\frac{2}{3}\) BE + \(\frac{2}{3}\) CF > BC ⇒ 2BE + 2CF > 3BC    ....(iv)

Similarly in ΔBGA, 

BG + GA > AB

⇒ \(\frac{2}{3}\)BE + \(\frac{2}{3}\)AD > AB ⇒ 2BE + 2AD > 3AB    …(v)

In ΔCGA, 

CG + GA > AC

⇒ \(\frac{2}{3}\) CF + \(\frac{2}{3}\)AD > AC

⇒ 2CF + 2AD > 3AC                        ....(vi)

Adding (iii), (iv) and (v), we get 

2BE + 2CF + 2BE + 2AD + 2CF + 2AD > 3BC + 3AB + 3AC 

⇒ 4(AD + BE + CF) > 3(AB + BC + AC)

⇒ 3(AB + BC + AC) < 4(AD + BE + CF) ⇒ III is true.