ATRACTOR

The are of the new polygon is \[\frac{1}{2}\left|\sum_{i=0}^{n-1} P'_i\times P'_{i+1}\right|\] where \((a,b)\times (c,d)=\det\left( \begin{array}{cc}a & b\\c & d\\\end{array}\right)=a d - b c\) for any vectors \((a,b)\) and \((c,d)\).
We have \[\begin{array}{ll}\sum_{i=0}^{n-1} P'_i\times P'_{i+1} & =\; \sum_{i=0}^{n-1} \frac{1}{2} (P_i+ P_{i+1}) \times \frac{1}{2} (P_{i+1}+ P_{i+2})\;=\\ & =\; \frac{1}{4}\sum_{i=0}^{n-1} (P_i\times P_{i+1} + P_i\times P_{i+2}+ P_{i+1}\times P_{i+2})\;=\\ & =\; \frac{1}{4}\sum_{i=0}^{n-1} P_i\times P_{i+1} + \frac{1}{4}\sum_{i=0}^{n-1} P_i\times P_{i+2}+ \frac{1}{4}\sum_{i=0}^{n-1} P_{i+1}\times P_{i+2}\;=\\ & =\; \frac{1}{4}\sum_{i=0}^{n-1} P_i\times P_{i+1} + \frac{1}{4}\sum_{i=0}^{n-1} P_i\times P_{i+2}+ \frac{1}{4}\sum_{i=0}^{n-1} P_i\times P_{i+1}\;=\\ & =\; \frac{1}{2}\sum_{i=0}^{n-1} P_i\times P_{i+1} + \frac{1}{4}\sum_{i=0}^{n-1} P_i\times P_{i+2} \end{array}\] If \(P\) is a triangle, we have: \[\begin{array}{ll}\sum_{i=0}^2 P_i\times P_{i+2} &=\;P_0\times P_2+P_1\times P_3+P_2\times P_4\;=\\ &=\;P_0\times P_2+P_1\times P_0+P_2\times P_1\;=\\ &=\;-(P_0\times P_1+P_1\times P_2+P_2\times P_0)\;=\\ &=\;-\sum_{i=0}^2 P'_i\times P'_{i+1} \end{array}\] so that \[\sum_{i=0}^2 P'_i\times P'_{i+1}\;=\;\frac{1}{2}\sum_{i=0}^2 P_i\times P_{i+1}-\frac{1}{4}\sum_{i=0}^2 P_i\times P_{i+1}\;= \;\frac{1}{4}\sum_{i=0}^2 P_i\times P_{i+1} \] and the area of the new triangle constrcuted by bisection is a quarter of the area of the initial triangle.
If \(P\) is a quadrilateral, we have \[\begin{array}{ll}\sum_{i=0}^3 P_i\times P_{i+2} &=\;P_0\times P_2+P_1\times P_3+P_2\times P_4+P_3\times P_5\;=\\ &=\;P_0\times P_2+P_1\times P_3+P_2\times P_0+P_3\times P_1\;=\;0 \end{array}\] so that \[\sum_{i=0}^3 P'_i\times P'_{i+1}\;=\;\frac{1}{2}\sum_{i=0}^3 P_i\times P_{i+1}\] and the area of the new quadrilateral obtained by bisection is half of the area of the initial quadrilateral.
Note: in the case of polygons with an even number of vertices (\(n=2m\) for some integer \(m\)) bigger than \(4\), we have \[\sum_{i=0}^{n-1} P_i\times P_{i+2}\;=\;\sum_{j=0}^{m-1} P_{2j}\times P_{2j+2} + \sum_{j=0}^{m-1} P_{2j+1}\times P_{2j+3}\] so that \(\frac{1}{2}\left|\sum_{j=0}^{m-1} P_{2j}\times P_{2j+2}\right|\) gives the area of the polygon with vertices \(P_{2j}\) (that is, with even index) and \(\frac{1}{2}\left|\sum_{j=0}^{m-1} P_{2j+1}\times P_{2j+3}\right|\) gives the area of he polygon with vertices \(P_{2j+1}\) (that is, with odd index). Hence, the area of the new polygon obtained by bisection is equal to the sum of half of the area of the initial polygon with a quarter of the areas of the polygons obtained by linking respectively the vertices of even index and the vertices of odd index of the initial polygon.