\({\text{ABCDEF}}\) is a hexagon. A circle lies inside the hexagon and touches each of the six sides.

Show that \({\text{AB}} + {\text{CD}} + {\text{EF}} = {\text{BC}} + {\text{DE}} + {\text{FA}}\).

     A1

the lengths of the two tangents from a point to a circle are equal     (R1)

so that

\({\text{AG}} = {\text{LA}}\)

\({\text{GB}} = {\text{BH}}\)

\({\text{CI}} = {\text{HC}}\)

\({\text{ID}} = {\text{DJ}}\)

\({\text{EK}} = {\text{JE}}\)

\({\text{KF}} = {\text{ FL}}\)     A1

adding,

\({\text{(AG}} + {\text{GB)}} + {\text{(CI}} + {\text{ID)}} + {\text{(EK}} + {\text{KF)}} = {\text{(BH}} + {\text{HC)}} + {\text{(DJ}} + {\text{JE)}} + {\text{(FL}} + {\text{LA)}}\)     M1A1

\({\text{AB}} + {\text{CD}} + {\text{EF}} = {\text{BC}} + {\text{DE}} + {\text{FA}}\)     AG

[5 marks]