Question

if AD, BE, CF are the medians of triangle ABC ,then prove that--

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

ii.3(AB+BC+CA) > 2(AD+BE+CF)

PLZ ANSWER IT GUYS........ I HAVE EXAM........​

Answer 1

(i)

BE + CF > (3/2) BC

=> 2(BE + CF) > 3BC

Similarly,

2(CF + AD) > 3CA

2(AD + BE) > 3AB

On combining all equations, we will get

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

(ii)

In ΔABD,

AB + BD > AD

In ΔBCE,

BC + CE > BE

In ΔACF,

CA + AF > CF

On adding 3 Equations, we will get

AB + BD + BC + CE + CA + AF > AD + BE + CF  ---- (iv)

D,E ad F are midpoints of BC,AC and AB.

=> BD = 1/2BC => CE = 1/2CA => AF = 1/2AB   ----- (v)

Then,

AB + 1/2BC + BC + 1/2CA + CA + 1/2AB > AD + BE + CF

=> 2AB + BC + 2BC + CA + 2CA + AB/2 > AD + BE + CF

=> 3AB + 3BC + 3CA > 2AD + 2BE + 2CF

=> 3(AB+BC+CA) > 2(AD+BE+CF)

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