# 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

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  + ( $\frac{1}{2}$ 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

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