How does one prove that the quadrilateral $DFPI$ is cyclic given that the greens angles are congruent? This is a solution to a larger problem, however I do not understand this step when they deduce from the green angles being the same size, that the quadrilaterial is cyclic. It doesn't seem too obvious to me.
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If two angles subtending the same segment are equal and both angle vertices lie in the same half-plane created by the line containing the segment, then the end points of the segment and the two angle vertices form a cyclic quadrilateral.
This can be easily proved. Draw the circle through the end points of the segment and one of the angle vertices. Then from the extended central angle theorem the other vertex lies on the same circle (if it would lie outside or inside the circle the corresponding angle would be less or larger than the first one, respectively).
In your example the segment is $IP$ and the angle vertices are $D$ and $F$.
The point $P$ is not any but a point located on the circumcircle circumscribed to the square. Try to draw another point outside the upper arc $FI$ of said circumference and you will not be able to have equality of the green angles.The point $B$ also belong to the circumcircle (not only points $F,P,I,D$).