If ABCDEF is a regular hexagon then prove that ab+ac+ad+ea+fa=4ab

Answer 1

I assume you mistyped the problem, it meant: ab+ac+ad+ea+fa=4ab where each quantity is a VECTOR.
first, note AB+BC=AC and
AD=AB+BC+CD and
EA (note direction, from E to A)=-(AB+BC+CD+DE) and
FA=-(EF+AE)
Does this help?

Answer 2

in a regular hexagone ABCDEF prove that AB + AC + AD + EA +FA = 4AB

Answer 3

In hexagon ABCDEF prove that AB + AC + AD + EA + FA = 4AB

Answer 4

If ABCDEF is a regular hexagon then Prove that:
AB+AC+AD+EA+FA=4AB