Q. 5

# The medians AD, BE and CF in a triangle ABC intersect at the point G. Prove that 4(AD + BE + CF) > 3(AB + BC + CA).

Answer :

Given that in ∆ABC , medians AD, BE and CF intersects at a point G.

As we know that the sum of any two sides is always greater than the third side in a triangle.

Here, G is the centroid of the ∆ABC

Now, in ∆ADB and ∆ADC,

AD+BD>AB………..(1)

AD+DC>AC…………(2)

In ∆BEC and ∆BEA,

BE+EC>BC………..(3)

BE+AE>AB………..(4)

In ∆CFA and ∆CFB,

CF+AF>AC………..(5)

CF+FB>BC………..(6)

Adding equation (1),(2),(3),(4),(5)and (6)

2AD+2BE+2CF+(BD+DC)+(EC+AE)+(AF+FB)>2AB+2BC+2AC

⇒ 2(AD+BE+CF)+BC+AC+AB>2(AB+BC+AC)

⇒ 2(AD+BE+CF)> 2(AB+BC+AC)-(BC+AC+AB)

⇒ 2(AD+BE+CF)> AB+BC+AC

Multiplying by 2 in the above equation

⇒ 4(AD+BE+CF)> 2(AB+BC+AC)

Hence, Proved.

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