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Why is the dividing set nonempty when a convex surface has Legendrian boundary?

I am an undergrad and curious why dividing set of a convex surface $\Sigma$ is nonempty when it has a Legendrian boundary?

I know you can use Stokes theorem to contrapositive the statement for closed case but I don't see why for Legendrian boundary case.

Thanks for your help!

p.s. Somebody had voted this question as "unclear or not useful"! I am just an undergrad who tries to understand something hard for me:( Such a discouraging move..