Two points inside a triangle x distance away from each other, at least one of it must be x distance away from one of the vertices of a triangle.

Question

Given a triangle $\triangle $ ABC. Suppose there are two points inside the triangle, and they are $p$ and $q$. Let $d(p,q)=x$, here d represents distance. How to prove this:

1. if $\min\{d(A,p), d(C,q)\}=d(A,p)$ then $d(A,q) \geq x$.

or

2. if $\min\{d(A,p), d(C,q)\}=d(C,q)$ then $d(C,p) \geq x$.

In the figure, I showed the first if statement.

PS: I am doing it by drawing a circle centred at q and of radius x.

Answer 1

Any triangle that fails must have all its vertices inside the blue area, but any triangle like this can't contain points $p$ and $q$.