Any hexagon in Pascal's triangle, whose vertices are 6 binomial coefficients surrounding any entry, has the property that:

• the product of non-adjacent vertices is constant.

• the greatest common divisor of non-adjacent vertices is constant.

Here is one such hexagon. As an example, we have that $4 \cdot 10 \cdot 15 = 6 \cdot 20 \cdot 5$, as well as $\gcd(4, 10, 15) = \gcd(6,20,5)$.

There is a quick proof here (pdf). The original proof should be in V. E. Hoggatt, Jr., & W. Hansell. "The Hidden Hexagon Squares." The Fibonacci Quarterly 9(1971):120, 133. but I cannot access it.

I am, however, intereseted in a purely combinatorial proof. I do not know how to approach this at all: I cannot see what the non-adjacent vertices represent and/or I do not know how to remodel their meaning. Can anyone help?

I have asked this question on math.se, I have not yet received a satisfactory anwser. (The anwser provided there, is somewhat combinatorial in spirit, but maybe one level down, from what I am looking for. It begins with "In symbols, the identity is..." and then only uses the in-or-out arguement to finish up. It would be perfect if someone shared a problem/application of this too.)

EDIT: To specify my question more closely, what I am looking for is some natural bijection between the two sets of triads that create the hexagon.

Thanks.