In △ABC, AD,BE and CF are medians to respective sides.
So,
BD=DC=12BC
AE=EC=12AC
AF=FB=12AB
By Pythagoras theorem in △AFC
AC2=CF2+AF2
AC2=CF2+(12AB)2
AC2=CF2+14AB2
4AC2=4CF2+AB2 ....... (i)
By Pythagoras theorem in △BCE
BC2=BE2+EC2
BC2=BE2+(12AC)2
BC2=BE2+14AC2
4BC2=4BE2+AC2 ........ (ii)
By Pythagoras theorem in △ADB
AB2=AD2+BD2
AB2=AD2+(12BC)2
AB2=AD2+14BC2
4AB2=4AD2+BC2 ....... (iii)
Adding (i),(ii) and (iii), we get
4AC2+4BC2+4AB2=4AD2+BC2+4BE2+AC2+4CF2+AB2
3AC2+3BC2+3AB2=4AD2+4BE2+4CF2
3(AC2+BC2+AB2)=4(AD2+BE2+CF2)
AD2+BE2+CF2BC2+CA2+AB2=34
∴(AD2+BE2+CF2):(BC2+CA2+AB2)=3:4
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