Ive been doing this problem for a quite some time. Can anyone help me? Thanks!

Here I'm going to present combinatorial method. Consider circle $\omega$ center at $A$ with radius $1$, and consider "red" point be arbitrary point on $\omega$ From point $A$, we draw lines perpendicular to $DC,CB,DB$ and denote their intersection with $\omega$ as show in figure below. We also denote intersection between $\omega$ with side of quadrilateral as show below For any "red" point $R$, we draw a line pass through vertices of quadrilateral which parallel or perpendicular to $AR$ And let their intersection form $3$ rectangles with diagonals "blue,green,pink" as show in figure below For rectangle with green diagonals, when we move red point, start from $G_1$ and go to $G_2$ and finish at $G_3$ We fixed $4$ vertices of the rectangle, $A_1,A_2,A_3,A_4$ At $G_1$, we see that the rectangle will coincide with $BD$, we get that $\frac{A_iA_{i+1}}{A_{i+1}A_{i+2}}=\infty$ for some $i\in \{ 1,2,3,4\}$ At $G_2$, we see that the rectangle will coincide with $AC$, we get that $\frac{A_iA_{i+1}}{A_{i+1}A_{i+2}}=0$ At $G_3$, we see that the rectangle will coincide with $BD$ again, we get that $\frac{A_iA_{i+1}}{A_{i+1}A_{i+2}}=\infty$ And note that $\frac{A_iA_{i+1}}{A_{i+1}A_{i+2}}$ is continuous when the red point move along arc of $\omega$ And when $\frac{A_iA_{i+1}}{A_{i+1}A_{i+2}} =1$ we get that the rectangle is square. So there exist at least two red points $R_1$ in minor arc $G_1G_2$ and $R_2$ in minor arc $G_2G_3$ that the rectangle is square. Similar method also work for rectangle with pink and blue diagonals, so we get at least $2+2+2=6$ "amazing" squares P.S. Note that since $AC$ is not perpendicular to $BC$ we get that $G_1,G_2,G_3$ are all distinct point And when two red points are in opposite direction, the rectangles obtain from those red points are same rectangles.

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thanks dark lord!