#### Answer

The proof is below.

#### Work Step by Step

We prove that the two results are the same. We will use a two-dimensional vector to prove this.
(Ai+Bj)((Ci+Dj)+(Ei+Fj))
We add the two first and then expand:
(Ai+Bj)((C+E)i+(D+F)j)
We use the definition of the dot product to find:
(AC+AE)+(BD+BF)
We now use the distributive property to prove this is an equivalent method:
(Ai+Bj)((Ci+Dj)+(Ei+Fj))
(Ai)(Ci+Dj)+Ai(Ei+Fj)+Bj(Ci+Dj)+Bj(Ei+Fj)
Simplifying gives:
AC+AE+BD+BF
This is the same result as above.