Prove that the area of the n -gon A_1A_2... A_n is…

Prove that the area of the $n$-gon $A_1A_2\ldots A_n$ is $P=\tfrac12\Big|[\overrightarrow{OA_1}\times\overrightarrow{OA_2}]+[\overrightarrow{OA_2}\times\overrightarrow{OA_3}]+\cdots+[\overrightarrow{OA_n}\times\overrightarrow{OA_1}]\Big|$ where $O$ is any point in the plane and $[\ \cdot\times\cdot\ ]$ is the signed area of the parallelogram (positive for a counter-clockwise turn).

Step 1: Fan triangulation

The triangles $OA_iA_{i+1}$ (with $A_{n+1}=A_1$) have signed areas $\tfrac12[\overrightarrow{OA_i}\times\overrightarrow{OA_{i+1}}]$. For $O$ inside a convex polygon, all of them have the same sign and exactly tile the polygon. For a general $O$ (or a non-convex simple polygon), the parts outside the polygon are counted once positively and once negatively and cancel. So the sum is the signed area of the polygon.

\[P=\tfrac12\Big|\sum_{i=1}^{n}[\overrightarrow{OA_i}\times\overrightarrow{OA_{i+1}}]\Big|\]

Step 2: Independence of $O$

For $\overrightarrow{O'A_i}=\overrightarrow{OA_i}-\vec t$: \[[\overrightarrow{O'A_i}\times\overrightarrow{O'A_{i+1}}]=[\overrightarrow{OA_i}\times\overrightarrow{OA_{i+1}}]+[\overrightarrow{OA_{i+1}}\times\vec t]-[\overrightarrow{OA_i}\times\vec t],\] and the last two terms telescope to $0$ around the closed polygon. $\checkmark$

Step 3: Special cases

• $n=3$, $O=A_1$: $P=\tfrac12|\overrightarrow{A_1A_2}\times\overrightarrow{A_1A_3}|$.
• $n=4$: for any $O$ (take $O$ as origin, $\vec a_i=\overrightarrow{OA_i}$),

  \[ (\vec a_3-\vec a_1)\times(\vec a_4-\vec a_2)=\vec a_1\times\vec a_2+\vec a_2\times\vec a_3+\vec a_3\times\vec a_4+\vec a_4\times\vec a_1, \]

  so $P=\tfrac12|\vec d_1\times\vec d_2|$ with the diagonals $\vec d_1=\overrightarrow{A_1A_3}$, $\vec d_2=\overrightarrow{A_2A_4}$ (compare Problem 100).