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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