IF AD, BE AND CF ARE MEDIANS OF ?ABC, PROVE THAT
3(AB^2+BC^2+CA^2) = 4(AD^2+ BE^2+ CF^2)?

Dear Student
Answer :

We can prove that if in ABC  AD , BE and CF are median and altitude to respective sides , So we get 

So,

BD  =  DC  = $\frac{1}{2}$ BC

AE  =  EC  =  $\frac{1}{2}$ AC
And
AF  =  FB  = $\frac{1}{2}$ AB

Now we apply Pythagoras theorem in  $∆$ AFC  , As :

AC²  = CF²  + AF² 

AC²  = CF²  + ( $\frac{1}{2}$ AB )²  ( As we know AF  = $\frac{1}{2}$ AB  )

AC²  = CF²  + $\frac{1}{4}$ AB² 
Taking LCM we get

4 AC²  =  4CF²  + AB²  ---------------------- ( 1 )

Now we apply Pythagoras theorem in  $∆$ BCE  , As :

BC²  =BE²  + EC² 

BC²  = BE²  + ( $\frac{1}{2}$ AC)²  ( As we know EC   = $\frac{1}{2}$ AC  )

BC²  = BE²  + $\frac{1}{4}$ AC² 
Taking LCM we get

4 BC²  =  4BE²  + AC²  ---------------------- ( 2 )
And
Now we apply Pythagoras theorem in  $∆$ ADB  , As :

AB²  = AD²  + BD² 

AB²  = AD²  + ( $\frac{1}{2}$ BC )²  ( As we know BD   = $\frac{1}{2}$ BC  )

AB²  = AD²  + $\frac{1}{4}$ BC² 
Taking LCM we get

4 AB²  =  4AD²  + BC²  ---------------------- ( 3 )

Now we add equation 1 , 2 and 3 ,and get

4 AC²  + 4 BC² + 4 AB²  =  4CF²  + AB² +   4BE²  + AC²  +  4AD²  + BC²   

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

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

Regards