Isn't that interesting? What a neat little problem.
The middle number between a and b = (a + b)/2
The middle number between b and c = (b + c) / 2
The middle number between c and d = (c + d)/2
The middle number between d and a = (d + a)/2
The sum of the numbers in the corners of
diagram 1 = a + b + c + d Do you agree.
Now look at diagram two. Start by putting a, b, c and d in the corners.
Now remove the brackets from what I found above.
diagram 2 = a/2 + b/2 + b/2 + c/2 + c/2 + d/2 + d/2 + a/2 Now collect all the like terms.
diagram 2 = a/2 + a/2 + b/2 + b/2 + c/2 + c/2 + d/2 + d/2
a/2 + a/2 = a does it not?
b/2 + b/2 = b
c/2 + c/2 = c
d/2 + d/2 = d
The sum of the middle numbers in diagram 2 = a + b + c + d
But that's the same sum as diagram 1, which was what you were asked to prove. You cannot come up with a counter example that will give a different result, at least in the positive integers.
The question provides you with room for a written answer. You are going to have to reproduce in some form what I've put in italics.
Thank you for posting.