Proof that a 5-gon cannot have fewer than 5 diagonal intersections

Question

I figured out that a convex $n$-gon has at most $n\choose{4}$ diagonal intersections.

But how can I show (proof) that a convex $5$-gon cannot have less that $5$ diagonal intersections?

Answer 1

I claim that the only way for a convex $n$-gon to have fewer than $\binom n4$ intersection points would be in the case where more than two diagonals intersect at one point. If you can justify that claim, and show that it is impossible to get more than two diagonals through a single point of intersection in a convex pentagon, you can prove that any convex pentagon must have exactly five diagonal intersections.