Answer :

(i)


As we know, if G is centroid, and centroid divides the median in 2:1


Then BE = BG and FC = CG


In


Δ BGC,


BG + GC > BC (sum of two sides greater than 3rd side)



2BE + 2FC > 3BC


Or


3BC < 2BE + 2FC ….(1)


Similarly,


3CA < 2CF + 2AD ….(2)


And 3AB < 2AD + 2BE ….(3)


Add (1), (2) and (3)


We get,


3 BC + 3CA + 3AB < 2BE + 2FC + 2CF + 2AD + 2AD + 2BE


3 (AB + BC + CA) = 4 BE + 4 FC + 4 AD


Or 4 (AD + BE + CF) > 3 (AB + BC + CA)


Hence proved.


(ii) In ΔACD


AC + CD > AD (sum of two sides greater than 3rd side)


Since, AD is median



……[1]


Similarly,


In ΔABE


……[2]


In ΔBFC


……[3]


Adding [1], [2] and [3], we get





Hence, Proved!


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