Show that in a group of 10 people (where any 2 are either friends or enemies), there are either 3 mutual friends or 4 mutual enemies$\dots$ Question:Show that in a group of 10 people (where any 2 are either friends or enemies), there are either 3 mutual friends or 4 mutual enemies, and there are either 3 mutual enemies or 4 mutual friends.

I'm really lost in this question. After hours of just thinking and having no progress, I looked at the answer key, and still lost. So I will give you the books answer key and point out where I'm lost.

Book's Solution:
By symmetry we need to prove only the first statement. Let $A$ be one of the people. Either $A$ has at least four friends, or $A$ has at least six enemies among the other nine people (since $3 + 5 < 9$). Suppose, in the first case, $B, C,D, E$ are all $A$'s friends. If any two of these are friends with each other, then we have found three mutual friends. Otherwise $\{B, C, D, E\}$ is a set of four mutual enemies of $A$. By Example 11, among $B, C, D, E, F, G$ there are either three mutual friends or three mutual enemies, who form with $A$, a set of four mutual enemies. 

One can ignore the last sentence, it's basically saying that in 6 people, there are either 3 mutual friends or 3 mutual enemies, which I completely understand already.

My Problem:
The proof actually make sense, except for the part where they said "Either A has at least 4 friends, or A has at least six enemies among the other nine people (since 3 + 5 < 9)", that 's the part thats bothering me. Where did they get the number $4$? I mean, by pigeon hole principle, $A$ has at lest $\lceil \dfrac{9}{2}\rceil = 5$ enemies or friends (not both of course). If you could help me out, I would appreciate it. I'm almost halfway the Discrete Mathematics text book and this is so far the most bizarre, and I would like to build some neural path and make this problem easier.
 A: $A$ has $9$ people who are each friends or enemies.  Let's write this as $$F_A+E_A=9.$$
So if $F_A \le 3$ then this implies $E_A \ge 6$.
But since $F_A$ is an integer, $F_A \not \le 3$ implies $F_A \ge 4$.
So we can conclude from  $$F_A \le 3 \text{ or } F_A \not \le 3$$ that $$E_A \ge 6 \text{ or } F_A \ge 4$$
A: Let us assume A has 9 relations with B,C,D,E,F,G,H,I,J (i.e rest of 9 people).
They can be either friends or enemies.
By pigeon hole principle, A has atleast ⌈9/2⌉=5 enemies or friends (not both of course).
Let us now consider A is friends with 5 people i.e with B,C,D,E,F. 
Now we know here any 2 are either friends or enemies.(B,C,D,E,F)
Now apply pigeon hole principle again. Consider them as 5 nodes where these are connected by 4 edges which are of value 'f' or 'e' i.e friends or enemies.
B---C---D---E---F.So we have ⌈4/2⌉=2.
So we have either 2 are enemies or 2 are friends vice versa. Now let us consider 2 are enemies and 2 are friends. 
So B--(f)--C--(f)--D-(e)--E--(e)--F.
(where 'e' is for enemy edge and 'f' is for friend edge).
All the nodes B,C,D,E,F are connected to A as well as 'f' friends.
From above we can see A,B,C,D are 4 mutual friends and D,E,F are 3 mutual enemies, which we have to proof.
(Drawing a graph for above might help. Similarly we can take 4 friends for A and prove it).
A: $A$ can't have fewer than four friends and fewer than six enemies, because with only three friends and five enemies, friends + enemies would be $5+3=8$.
