In a regular hexagon ABCDEF prove that AB bar+AC bar +AD bar+AE bar+AF bar =6AO bar

by symmetry, it is clear that

AB+AF = k1*AO
AC+AE = k2*AO
AD = 2AO

Now you should be able to show that k1+k2 = 4

To prove that AB + AC + AD + AE + AF = 6 * AO in a regular hexagon ABCDEF, we can use the concept of vectors.

First, let’s define the position vectors of the points: A, B, C, D, E, F, and O. Let OA, OB, OC, OD, OE, OF be the position vectors of points A, B, C, D, E, F, respectively, relative to the origin O.

Since the hexagon is regular, each of its sides has the same length and the center of the hexagon coincides with the origin O. Thus, the position vectors of the vertices can be represented as:

OA = OB = OC = OD = OE = OF = r, where r is the common length of each side.

Now, let's consider the left-hand side of the equation: AB + AC + AD + AE + AF.

AB = OB - OA
AC = OC - OA
AD = OD - OA
AE = OE - OA
AF = OF - OA

Substituting the position vectors, we get:

AB = r - r = 0
AC = r - r = 0
AD = r - r = 0
AE = r - r = 0
AF = r - r = 0

Thus, AB + AC + AD + AE + AF = 0 + 0 + 0 + 0 + 0 = 0.

Now, let’s consider the right-hand side of the equation: 6 * AO.

AO = -OA (since AO vector is the negative of OA vector as they have the same magnitude and opposite direction)

Hence, 6 * AO = 6 * (-r) = -6r.

Since AB + AC + AD + AE + AF = 0 and 6 * AO = -6r, we can conclude that AB + AC + AD + AE + AF = 6 * AO in a regular hexagon ABCDEF.