Answer: B
AD < AB + BD BE < BC + EC CF < AC + AF ∴ AD + BE + CF < AB + BC + AC + \(\frac{1}{2}\)(BC + AC + AB) ⇒ AD + BE + CF < \(\frac{3}{2}\)(BC + AC + AB) ∴ \(x < \frac{3}{2}y\) That is, \(x < y\)
AD < AB + BD BE < BC + EC CF < AC + AF ∴ AD + BE + CF < AB + BC + AC + \(\frac{1}{2}\)(BC + AC + AB) ⇒ AD + BE + CF < \(\frac{3}{2}\)(BC + AC + AB) ∴ \(x < \frac{3}{2}y\) That is, \(x < y\)