Question

ABCDEF is a regular hexagon. Show that `bar"AB" + bar"AC" + bar"AD" + bar"AE" + bar"AF" = 6bar"AO"`, where O is the centre of the hexagon.

Answer 1

ABCDEF is a regular hexagon.

∴ `bar"AB" = bar"ED"  "and"  bar"AF" = bar"CD"`

∴ by the triangle law of addition of vectors,

`bar"AC" + bar"AF" = bar"AC" + bar"CD" = bar"AD"`

`bar"AE" + bar"AB" = bar"AE" + bar"ED" = bar"AD"`

∴ LHS = `bar"AB" +bar"AC" + bar"AD" + bar"AE" + bar"AF"`

`= bar"AD" + (bar"AC" + bar"AF") + (bar"AE" + bar"AB")`

`= bar"AD" + bar"AD" + bar"AD"`

`= 3bar"AD" = 3(2bar"AO")`    ....[O is midpoint of AD]

`= 6 bar"AO"`.