Question

AD, BE and CF are medians of the triangle ABC, then prove that i). 4(AD+BE+CF)>3(AB+BC+CA)
Ii). 3(AB+BC+CA)>2(AD+BE+CF)

Answer 1

BD = DC = 12 BC

AE = EC = 12 AC
And
AF = FB = 12 AB

Now we apply Pythagoras theorem in ∆ AFC , As :

AC2 = CF2 + AF2

AC2 = CF2 + ( 12 AB )2 ( As we know AF = 12 AB )

AC2 = CF2 + 14 AB2
Taking LCM we get

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

Now we apply Pythagoras theorem in ∆ BCE , As :

BC2 =BE2 + EC2

BC2 = BE2 + ( 12 AC)2 ( As we know EC = 12 AC )

BC2 = BE2 + 14 AC2
Taking LCM we get

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

AB2 = AD2 + BD2

AB2 = AD2 + ( 12 BC )2 ( As we know BD = 12 BC )

AB2 = AD2 + 14 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 )

Answer 2

Answer:

BD = DC = 12 BC

AE = EC = 12 AC

And

AF = FB = 12 AB

Now we apply Pythagoras theorem in ∆ AFC , As :

AC2 = CF2 + AF2

AC2 = CF2 + ( 12 AB )2 ( As we know AF = 12 AB )

AC2 = CF2 + 14 AB2

Taking LCM we get

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

Now we apply Pythagoras theorem in ∆ BCE , As :

BC2 =BE2 + EC2

BC2 = BE2 + ( 12 AC)2 ( As we know EC = 12 AC )

BC2 = BE2 + 14 AC2

Taking LCM we get

4 BC2 = 4BE2 + AC2 ---------------------- ( 2 )

And

Now we apply Pythagoras theorem in ∆ ADB , As :

AB2 = AD2 + BD2

AB2 = AD2 + ( 12 BC )2 ( As we know BD = 12 BC )

AB2 = AD2 + 14 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 )