Question

prove that in a triangle the sum of the squares of the side is 4 times the sum of the squares of the median

Answer 1

according to appollonius theorem  the sum of the square of 2 sides  of a triangle is equal to twice the square of the median on the third side  plus half the square of the third side.

AB^2+AC^2=2BD^2+2AD^2
=2*(I/2 BC)^2+2AD^2
=1/2BC^2+2AD^2
THEREFORE 2AB^2+2AD^2=BC^2+4AD {1}
SIMILARLY
2 AB^2 +2BC^2=AC^2+4BE^2 {2}
2BC^2+2AC^2=AB^2+4CF^2 {3}
 ADDING 1 ,2 AND 3
4AB^2+4BC^2+4AC^2=AB^2+BC^2+4AD^2+4BE^2+4CF^2
3(AB^2+BC^2+AC^2)=4(AD^2+BE^2+CF^2)

Answer 2

see diagram.

AD is the median on to BC. BE and CF are medians onto AC and AB respectively.  Draw a normal AG from A on to BC.

AB² = AG² + BG²
     = AD² - DG² +  BG²
     = AD² + (BG + DG) (BG - DG)
     = AD² + BD (BD -  DG - DG)
     = AD² + BD² -  2 * BD * DG      ---- (1)

AC² = AG² + GC² 
       = AD² - GD² + GC²
       = AD² + (GC - GD) (GC + GD)
       = AD² + DC * (2 GD + DC)
     = AD² + DC² + 2 DC * GD  --- (2)

now we obtain the Apolloneous theorem, as below.  as  DC = BD = 1/2 BC,

      AC² + AB² = 2 AD² + 2 BD²   = 2 AD² + 1/2 BC²   --- (3)

similarly we can prove that:
   BC² + BA² = 2 BE² + 2 AE² = 2 BE² + 1/2 AC²        --- (4)
   CB² + CA² = 2 CF² + 2 AF² = 2 CF² + 1/2 AB²          --- (5)

add  (3), (4) and (5) to get:
 
     2 (AB² + AC² + BC²) = 2(AD² + BE² + CF²) + 1/2 * (AB²+ AC²+BC²)

     3 (AB² + AC² + BC²)  = 4 ( AD² + BE² + CF²)