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This is a theorem of Cramer. See here

For the quadrilateral case the quickest proof is using Brahmagupta's formula

$$Area=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2 \theta}$$ where $a,b,c,d$ are the sides, $s$ is the half perimeter and $\theta$ is half the sum of opposite angles.

Edit: I wonder if this argument works: Pick four consecutive vertices and move the linkage made of these four vertices till it's cyclic. There will be a subsequence of the polygons we get after such operations which converges, by the Weierstrass theorem. In the limit the polygon will be cyclic otherwise you can find four consecutive vertices not on a circle and increase the area again.

This is a theorem of Cramer. See "On the impossibility of one rule-and-compass construction" by Vladimir Jankovic.

For the quadrilateral case the quickest proof is using Brahmagupta's formula

$$Area=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2 \theta}$$ where $a,b,c,d$ are the sides, $s$ is the half perimeter and $\theta$ is half the sum of opposite angles.

Edit: I wonder if this argument works: Pick four consecutive vertices and move the linkage made of these four vertices till it's cyclic. There will be a subsequence of the polygons we get after such operations which converges, by the Weierstrass theorem. In the limit the polygon will be cyclic otherwise you can find four consecutive vertices not on a circle and increase the area again.

This is a theorem of Cramer. See here

For the quadrilateral case the quickest proof is using Brahmagupta's formula

$$Area=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2 \theta}$$ where $a,b,c,d$ are the sides, $s$ is the half perimeter and $\theta$ is half the sum of opposite angles.

This is a theorem of Cramer. See here

For the quadrilateral case the quickest proof is using Brahmagupta's formula

$$Area=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2 \theta}$$ where $a,b,c,d$ are the sides, $s$ is the half perimeter and $\theta$ is half the sum of opposite angles.

Edit: I wonder if this argument works: Pick four consecutive vertices and move the linkage made of these four vertices till it's cyclic. There will be a subsequence of the polygons we get after such operations which converges, by the Weierstrass theorem. In the limit the polygon will be cyclic otherwise you can find four consecutive vertices not on a circle and increase the area again.

This is a theorem of Cramer. See here

This is a theorem of Cramer. See here

For the quadrilateral case the quickest proof is using Brahmagupta's formula

$$Area=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2 \theta}$$ where $a,b,c,d$ are the sides, $s$ is the half perimeter and $\theta$ is half the sum of opposite angles.