Solution:

Begin by letting $A$ be any person at the party, and note that $A$ must be either friends or strangers with at least three of the others, for otherwise there would only be at most $2+2=4$ other people at the party. Because of the symmetry between friends and strangers, we can assume that $A$ has three friends, call them $B, C$, and $D$. Now if any two of these three are friends, then they together with $A$ are the desired triple. Otherwise, $B, C$, and $D$ are all mutual strangers, so they themselves furnish the desired triple.