# 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

*print*Print*list*Cite

Here we use the relation between the lengths of the sides and that of the sides. Therefore we get:

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

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

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

Adding the three we get:

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

adding and subtracting the common terms

= (3/2) AC^2 + (3/2) BC^2 + (3/2) AB^2

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

**Therefore we get the result:**

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

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)