If in a triangle ABC ,AD,BE and CF are medians , then prove that: 3(AB^2+BC^2+CA^2) = 4(AD^2+BE^2+CF^2)step by step proving

Let us have the triangle
ABC.

Let D , E and F be the mid points of BC, CA and
AB.

Consider the triangle
ADB:

AB^2 = AD^2+ BD^2 - 2AD*DB cos ADB . Due to cosine
rule.

AB^2 = AD^2 +(BC/2)^2 - AD*BC* cos(ADB)....(1), as D
is mid point of BC.

Similarly  if we conseder trangle ADC,
we get:

AC^2 =AD^2+(1/2 B/2)^2 -
AD*BC*cos(ADC)......(2)

(1)+(2):

AB^2+BC^2
= 2AD^2 + ((BC)^2)/2 - AD*BC{cosADB+cosADC).....(3).

But 
the angles ADB and ADC are supplementary angles. So cosADB +cosADC = 0. Therefore eq (3)
becomes:

AB^2+AC^2 = 2AD^2 +(1/2)BC^2....
(4)

Similarly we can show by considering triangle BEC and
BED that

BC^2+BA^2 = 2BE^2
+(1/2)CA^2.........(5)

Similarly we can show
that

CA^2+AB^2 = 2CF^2
+(1/2)AB^2................(4)

(4)+(5)+(6):

2(AB^2+BC^2+CA^2)
=2(AD^2+BE^2+CF^2) +(1/2) {AB^2+BC^2+CA^2).

Subtract (1/2)
(AB^2+BC^2+CA^2) we get:

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

Multiply by
2:

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