From Boundary of Polygon is Jordan Curve, it follows that the boundary of $\triangle$ is equal to the image of a Jordan curve, so $\Int \triangle$ is well-defined. === Interior is Subset === Let $q \in \Int \triangle$. Let $i \in \set {1, 2, 3}$, and put $j = i \bmod 3 + 1$, $k = \paren {i + 1} \bmod 3 + 1$. Let $S_i$ be the side of $\triangle$ that is adjacent to $A_i$ and $A_j$, let $S_j$ be the side adjacent to $A_j$ and $A_k$, and let $S_k$ be the side adjacent to $A_k$ and $A_i$. Define $\mathbf v = A_j - A_i \in \R^2$, $\mathbf w = A_k - A_i \in \R^2$, and define $\mathbf u = \mathbf v - \mathbf w$. Define two rays $\LL = \set {q + s \mathbf u: s \in \R_{\ge 0} }$, and $\LL' = \set {q + s' \paren {-\mathbf u} : s' \in \R_{\ge 0} }$. As both rays are parallel to $S_j$, $\LL$ or $\LL'$ can only cross $S_j$ if $S_i$ and $S_k$ lie on opposite sides of $S_j$. This would imply that there is a non-convex angle in $\triangle$. As Sum of Angles of Triangle Equals Two Right Angles shows, this is impossible, as a non-convex angle is larger than two right angles. Then, neither $\LL$ nor $\LL'$ crosses $S_j$. As $\LL \cup \LL'$ is a straight line, and $\LL \cap \LL' = \set q$, it follows that $\LL$ and $\LL'$ cannot both intersect the same side. Then Jordan Polygon Interior and Exterior Criterion shows that $\LL$ and $\LL'$ each crosses one of the sides $S_i$ and $S_k$. When we denote the intersection points $p_1, p_2 \in \R^2$, we have: :$p_1 = A_i + r_1 \mathbf v = q + r \paren {\mathbf v - \mathbf w}$ :$p_2 = A_i + r_2 \mathbf w = q + r' \paren {\mathbf v - \mathbf w}$ for some $r_1, r_2 \in \openint 0 1$, where either $r, -r' \in \R_{>0}$ or $-r, r' \in \R_{>0}$. Subtracting the two equations gives: :$r_1 \mathbf v - r_2 \mathbf w = \paren {r - r'} \paren {\mathbf v - \mathbf w}$ which can be rearranged as: :$\paren {r_1 + r' - r} \mathbf v - \paren {r_2 + r' - r} \mathbf w$ As $\mathbf v$ and $\mathbf w$ are direction vectors for the adjacent sides $S_i$ and $S_k$, they cannot be parallel, so $v$ and $w$ are linearly independent. It follows that $0 = r_1 + r' - r = r_2 + r' - r$, so $r_1 = r_2$. Adding the two equations gives: :$2 A_i + r_1 \mathbf v + r_2 \mathbf w = 2 q + \paren {r + r'} \paren {\mathbf v - \mathbf w}$ which can be rearranged to give an expression for $q$: {{begin-eqn}} {{eqn | l = q       | r = A_i + \paren {\dfrac 1 2 r_1 - \dfrac 1 2 r - \dfrac 1 2 r'} \mathbf v + \paren {\dfrac 1 2 r_1 + \dfrac 1 2 r + \dfrac 1 2 r'} \mathbf w       | c = as $r_1 = r_2$ }} {{eqn | r = A_i + \paren {\dfrac 1 2 - \dfrac{r + r'} {2 r_1} } r_1 \mathbf v + \paren {1 - \paren {\dfrac 1 2 - \dfrac {r + r'} {2 r_1} } } r_1 \mathbf w }} {{end-eqn}} This shows that $q \in U_i$. As $i \in \set {1, 2, 3}$ was arbitrary, it follows that $\displaystyle q \in \bigcap_{i \mathop = 1}^3 U_i$. {{qed|lemma}} === Interior is Superset === Let $\displaystyle q \in \bigcap_{i \mathop = 1}^3 U_i$. For $i \in \set {1, 2, 3}$, define $j, k \in \set {1, 2, 3}$, the sides $S_i, S_j, S_k$ of $\triangle$, and their direction vectors  $\mathbf v, \mathbf u, \mathbf w$ as in the section above. As $q \in U_i$, it follows that $q = A_i + st \mathbf v + t \paern {1 - s} t \mathbf w$ for some $s \in \openint 0 1$ and $t \in \R_{>0}$. Let $\LL = \set {q + r \paren {-\mathbf v}: r \in \R_{\ge 0} }$ be a ray with start point $q$. If $\LL$ crosses the side $S_i$ that $\LL$ is parallel to, then the intersection point is: :$A_i + r_1 \mathbf v = q - r \mathbf v$ for some $r_1 \in \closedint 0 1, r \in \R_{\ge 0}$, which can be rearranged as: :$q = A_i + \paren {r_1 - r} \mathbf v + 0 \mathbf w$ As $\mathbf v$ and $\mathbf w$ are linearly independent, this implies $\paren {1 - s} t = 0$. Then either $t = 0$ or $s = 1$, which is impossible, so $\LL$ does not intersect $S_i$. However, we will show that $\LL$ crosses $S_k$, as the intersection point is: {{begin-eqn}} {{eqn |l= A_i + r_2 \mathbf w       |r= q - r \mathbf v }} {{eqn |r= A_i + st \mathbf v + t \paren {1 - s} t \mathbf w - r \mathbf v }} {{end-eqn}} for some $r_2 \in \closedint 0 1$, $r \in \R_{>0}$, which implies: :$\mathbf 0 = \paren {s t - r} \mathbf v + \paren {t \paren {1 - s} - r_2} \mathbf w$ As $v$ and $w$ are linearly independent, we have $0 = st - r$ and $0 = t \paren {1 - s} - r_2$. Then $r = st \in R_{>0}$, and $r_2 = t \paren {1 - s} \in \R_{>0}$. We must show that $r_2 < 1$. As $q \in U_k$, we have $q = A_k + s't' \paren {-\mathbf w} + t' \paren {1 - s'} \mathbf u$ for some $s' \in \openint 0 1$, $t' \in \R_{>0}$. As $A_k = A_i + \mathbf w$, we have: :$q - r \mathbf v = A_i + \mathbf w - s't' \mathbf w + t' \\paren {1 - s'} \mathbf v - t' \paren {1 - s'} \mathbf w - r \mathbf v$ As $q - r \mathbf v = A_i + r_2 \mathbf w$, we rearrange the equality to obtain: :$\mathbf 0 = \paren {t' \paren {1 - s'} - r} \mathbf v + \paren {1 - r_2 - s't' - t' \paren {1 - s'} } \mathbf w$ As $\mathbf v$ and $\mathbf w$ are linearly independent, this gives two equations: :$0 = t' \paren {1 - s'} - r$ :$0 = 1 - r_2 - s't' - t' \paren {1 - s'}$ Adding these equations gives: :$s't' + r = 1 - r_2$ As $s't' > 0$ and $r > 0$, it follows that $r_2 < 1$. Then, we have shown that $\LL$ crosses $S_k$. Let $\LL' = \set {q + r \mathbf v : r \in \R_{\ge 0} }$ be a ray with start point $q$. An argument similar to the one above shows that $\LL'$ crosses the side $S_j$. As $\LL \cup \LL'$ is a straight line, $\LL$ and $\LL'$ cannot both intersect the same side. It follows that $\LL$ has one crossing of the boundary of $\triangle$, so the parity of $q$ is $\map {\mathrm {par} } q = 1$. From Jordan Polygon Interior and Exterior Criterion, it follows that $q \in \Int \triangle$. {{qed|lemma}} The result now follows by definition of set equality. {{qed}} Category:Topology 4iuiu6w17z89omsjuyr48ykh9htwp6p