Why do squares with center P satisfy the given equations in Putnam A1?

[john562]

Question:
Let f be a real-valued function on the plane such that
for every square ABCD in the plane, f(A) + f(B) +
f(C) + f(D) = 0. Does it follow that f(P ) = 0 for all
points P in the plane?

Answer:
Yes, it does follow. Let P be any point in the plane. Let
ABCD be any square with center P . Let E; F; G; H
be the midpoints of the segments AB; BC; CD; DA,
respectively. The function f must satisfy the equations
0 = f(A) + f(B) + f(C) + f(D)
0 = f(E) + f(F ) + f(G) + f(H)
0 = f(A) + f(E) + f(P ) + f(H)
0 = f(B) + f(F ) + f(P ) + f(E)
0 = f(C) + f(G) + f(P ) + f(F )
0 = f(D) + f(H) + f(P ) + f(G):
If we add the last four equations, then subtract the first
equation and twice the second equation, we obtain 0 =
4f(P ), whence f(P ) = 0.

Comments:
I don't understand why
0 = f(A) + f(E) + f(P ) + f(H)
0 = f(B) + f(F ) + f(P ) + f(E)
0 = f(C) + f(G) + f(P ) + f(F )
0 = f(D) + f(H) + f(P ) + f(G)?

 

[Yuqing]

Those are smaller squares within the larger square ABCD. To be specific, they are the four quarters of ABCD

 

[john562]

I don't understand how adding them equals zero.

 

[mxbob468]

what is this? like the easiest putnam problem ever?