Triangle inside a simply connected open subset of the complex plane

Question

Let $U$ be a connected open subset of the complex plane. Suppose $U$ is simply connected, i.e. its fundamental group is trivial. Let $T$ be a triangle whose boundary is contained in $U$. It is intuitively obvious that the interior of $T$ is contained in $U$. How can we prove this? If it is too easy, how about if $T$ is a polygon?

Answer 1

Say $Ind(\gamma, p)$ is the index (winding number) of the closed curve $\gamma$ about $p$. What you want is a special case of

Proposition If $U$ is simply connected, $\gamma$ is a closed curve in $U$, and $p\in\Bbb C\setminus U$ then $Ind(\gamma,p)=0$.

Proof. Since $U$ is simply connected, there is a continuous family $(\gamma_t)_{0\le t\le 1}$ of closed curves in $U$ such that $\gamma_0=\gamma$ and $\gamma_1$ is constant. Now $Ind(\gamma_t,p)$ takes only integer values and depends continuously on $t$, so it is constant; hence $Ind(\gamma,p)=Ind(\gamma_1,p)=0$.