In triangleABC AD,BE,CF are the medians Prove that 3(AB^2+BC^2+AC^2)=4(AD^2+BE^2+CF^2)

Take the triangle ABC.

We
know that AD, BE and CF are the medians. Therefore D , E and F be the mid points of BC,
CA and AB resp.

We first consider the triangle
ADB:

Now according to the cosine rule
:

AB^2 = AD^2+ BD^2 - 2AD*DB cos
ADB .

as D bisects
BC

=> AB^2 = AD^2 +(BC/2)^2 - 2*AD*(BC/2)*cos ADB
...(1)

Similarly, for triangle ADC, we
get:

AC^2 = AD^2 + DC^2 - AD*DC*cos
ADC

=> AC^2 = AD^2 + (BC/2)^2 - 2*AD*(BC/2)*cos ADC
...(2)

Adding (1) and
(2)

=> AB^2 + BC^2 = 2AD^2 + 2*(BC/2)^2) -
AD*BC*[cos ADB + cos ADC]

Now as ADB and ADC are are
supplementary angles. So cos ADB + cos ADC = 0.

=>
AB^2 + AC^2 = 2AD^2 + (1/2)*BC^2

Similar results can be
obtained for the other medians. Therefore we have :

BC^2 +
BA^2 = 2BE^2 +(1/2)*CA^2

CA^2 + AB^2 = 2CF^2
+(1/2)*AB^2

AB^2 + AC^2 = 2AD^2 +
(1/2)*BC^2

Adding the
three

=> 2(AB^2 + BC^2 + CA^2) = 2(AD^2 + BE^2 +
CF^2) + (1/2)*(AB^2+BC^2+CA^2)

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

=> 3(AB^2+BC^2+CA^2)
= 4(AD^2+BE^2+CF^2)

Therefore we get

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