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

Hello students
Divide thehexagoninto 4 triangles by joiningAC,AD,AE. Now,AF= CD, because its aregular hexagon, therefore it has equal and parallel opposide sides. Add one moreADboth the sides and v get the desired result. therefore,AB+AC+AD+AE+AF=3(2AD)=6AO.

Hope it helps
thankyou