Let position vector of A,B,C and D be respectively a,b,c and d ‌∴‌‌AB=b−a,‌‌CB=b−c ‌CD=d−c,‌‌AD=d−a ‌∴AB+CB+CD+AD ‌=2b+2d−2a−2c ‌=2(b+d−a−c) And let position vector of E and F are e and f ∴‌e=‌
a+c
2
⇒a+c=2e ‌ And ‌‌f=‌
b+d
2
⇒b+d=2f ∴ From Eq. (i), we get ‌AB+CB+CD+AD=2(2f−2c) ‌=4(f−e)=4EF