# 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

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

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

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)

neela | High School Teacher | (Level 3) Valedictorian

Posted on

Let us have the triangle ABC.

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

Similarly  if we conseder trangle ADC, we get:

(1)+(2):

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):