ABCDEF is a regular hexagon.if AB=a BC=b,express AC ,AD, AND AE In terms of a and b. Also prove that AB +BC+ CD+ DE +EF +FA=0

To express AC, AD, and AE in terms of a and b, we can use the properties of a regular hexagon.

1. AC: In a regular hexagon, any two consecutive sides are equal. Therefore, AC is equal to BC, which is b.

2. AD: In a regular hexagon, if we draw a line from any vertex to the center, it bisects the angle formed by the two adjacent sides. This means that triangle ACD is an isosceles triangle, and AD is equal to CD. So AD = CD = b.

3. AE: Again, in a regular hexagon, if we draw a line from any vertex to the center, it divides the hexagon into two congruent triangles. Triangle AEF is congruent to triangle ABC, so AE is equal to AB, which is a.

Thus, we have:
AC = b
AD = CD = b
AE = a

Now, let's prove that AB + BC + CD + DE + EF + FA = 0.

In a regular hexagon, every pair of opposite sides has the same length. Therefore, AB = DE, BC = EF, and CD = FA. So, we can rewrite the expression as:

AB + BC + CD + DE + EF + FA = AB + BC + CD + AB + BC + CD
= 2(AB + BC + CD)

Since AB = CD and BC = EF, the expression simplifies to:

2(AB + BC + CD) = 2(AB + BC + AB)
= 4(AB + BC)

Now, recall that in a regular hexagon, the sum of the lengths of all sides is zero. This means that AB + BC + CD + DE + EF + FA = 0.

Therefore, the expression AB + BC + CD + DE + EF + FA evaluates to zero based on the properties of a regular hexagon.

AC = AB+BC = a+b

and so on.

Of course they sum to zero, since that is a closed path.