By Pascal theorem for the degenerate cyclic hexagon $EEDBAC,$ the intersections $P \equiv EE \cap AB,$ $ED \cap AC$ and $Q \equiv DB \cap CE$ are collinear, but $ED \cap AC$ is the infinity point of the parallel lines $ ED,AC$ $\Longrightarrow$ $ED \parallel AC \parallel PQ.$ Since $PA \parallel QC,$ it follows that $PACQ$ is a parallelogram $\Longrightarrow$ Segments $AC$ and $PQ$ are congruent and parallel.

we know that $AE=BC$ and $AE=DC$ $\longrightarrow$ $DC=BC$ so $BDC=CBD$ $\longrightarrow$ $CQ=$ ${DC}^{2}$ $/$ $EC$ other wise from 2 triangle PAE and AEC we know that $PA=$ ${AE}^{2}$ $/$ $EC$ so $PA=QC$ so $PAQC$ is a parallelogram! im so slothy excuse me for my bad write if you not catch say to explain it!!

Let’s forget that circle is called $c$ and let it be $\omega$. We can take it as unit circle centered at $0$ on complex plane. Denote with small letter $x$ complex number representing point $X$. Since $\omega$ is the unit circle $$(AB\parallel EC\wedge AC\parallel ED)\iff \left(e=\frac{ab}{c}\wedge d=\frac{c^2}{b}\right).$$From intersection formula $$q=\frac{bd(c+e)-ce(b+d)}{bd-ce}=\frac{c(c^2+ab)-a(b^2+c^2)}{c^2-ab},$$$$p=\frac{e^2(a+b)-2abe}{e^2-ab}=\frac{ab(a+b-2c)}{ab-c^2}.$$Thus $$p-q=\frac{c^3+a^2b-abc-ac^2}{ab-c^2}=a-c.$$In particular $$|p-q|=|a-c|$$QED.