ABCDEF is a regular hexagon, prove that AB+AC+AD+AE+AF=3AD

							Hi
Divide the hexagon into 4 triangles by joining AC, AD, AE .
Considering triangle ACD, v get AD= AC + CD
for Tri ADE , AD= AE + ED
Add the two eq=>
2AD= AC + CD + AE + ED
Now, AF = CD, because its a regular hexagon, therefore it has equal and parallel opposide sides.
Similarly , ED= AB
Substitute in the eq :
2AD= AB+ AC + AE + AF
Add one more AD both the sides and v get the desired result.
						

							From vector properties,
AB=ED,
AE+AB=AD,        ...(1)
Similarly,
AF=CD,
AC+AF=AD       ...(2)
From (1), (2)
AF+AC+AD+AE+AB=AD+AD+AD=3AD
						

							Here,To prove: AB+AC+AD+AE+AF=3ADNow,,  LHS= AB+AC+AD+AE+AF       = AB+AE+AC+AF+AD       = AE+ED+AC+CD+AD ( opposite sides of regular hexagon are equal So AB=ED & AF=CD )       = AD+AD+AD       =3AD #proved              =
						

							given: ABCDEF is a regular hexagon with center O.AB+AC+AD+AE+AF=AB+(AC+AF)+AD+AE=AB+(AC+CD)+AD+AE [since AF=CD]=AB+AD+AD+AE=2AD+(AB+AE)=2AD+(ED+AE)  [since AB=ED=2AD+AD=3AD=3*(2AO)   [since O is the center and AO=OD=6AOhope this helps you
						

							
AB+AC+AD+AE+AF
=(AB+AE)+(AC+AF)+AD
=AD+AD+AD+=3AD=3(2AO)=6AO
therefore,AB+AC+AD+AE+AF=3(2AD)=6AO.
						

							
As we know all side of  the regular hexagon are equal 
AB+AC+AD+AE+AF=3ADNow,, LHS= AB+AC+AD+AE+AF = AB+AE+AC+AF+AD = AE+ED+AC+CD+AD ( opposite sides of regular hexagon are equal So AB=ED & AF=CD ) = AD+AD+AD =3AD #prove