(d) In a regular hexagon ABCDEF, We know that, AB+BC+CD=AD=2BC
∴‌‌AB+CD=BC‌‌.....(i) Similarly, as ‌‌AF=CD‌‌.......(ii) and as we know that CD+DE+EF=CF=2DF ⇒‌‌CD+EF=DE‌‌........(iii) from Eqs. (i) and (iii), we get AB+CD+CD+EF=BC+DE ⇒‌‌AB+AF+CD+EF=BC+DE {from eq (ii)}