An ellipse with foci $E,F$ is tangent to four sides of quadrilateral $ABCD$, then
$$\tag{eq1} AB⋅EC⋅DF=FA⋅BE⋅CD$$
The order of points in the equation is interesting: $ABECDF$ is a cyclic shift of $FABECD$.
$$\tag{eq2}\frac{CE⋅CF}{CB⋅CD}+\frac{DE⋅DF}{DA⋅DC}=1$$
These identities are needed in my proof of another identity.
The identity holds for the complex number $a,b,c,d,e,f$ corresponding to points $A,B,C,D,E,F$: $$ \frac{(a-b) (e-c) (d-f)}{(f-a) (b-e) (c-d)}=-1 $$
Taking $\arg$, we get $\arg\frac{a-b}{a-f}+\arg\frac{e-c}{e-b}+\arg\frac{d-f}{d-c}\equiv0\pmod{2\pi}$.
Taking $|\cdot|$, we get the desired identity $\rm(eq1)$.
By Corollary 2.1 in a blogpost, isogonal conjugation is just $P\mapsto\frac1P$, in a suitable coordinate system!
In our setting, Corollary 2.1 implies the complex identity $$\tag1\label1 (m-a)(m-c)=(m-e)(m-f) $$ Miquel point of $a,b,c,d$ is defined by $$\tag2\label2 (m-a)(m-c)=(m-b)(m-d)$$ By scaling the coordinates, set $(m-a)(m-c)=(m-b)(m-d)=1$.
By translating the coordinates, set $m=0$, then \eqref{1} and \eqref{2} becomes $$ac=bd=ef=1$$ Plugging in our identity $$\frac{(a-b) (e-c) (d-f)}{(f-a) (b-e) (c-d)}+1=\frac{(a-b) (e-a^{-1}) (b^{-1}-e^{-1})}{(e^{-1}-a) (b-e) (a^{-1}-b^{-1})}+1=0$$ we are done!
This choice of coordinates greatly simplifies our computation. We can use the same method to prove identity $\rm(eq2)$: $$\tag3\label3 \frac{(c-e) (c-f)}{(c-b) (c-d)}+\frac{(d-e) (d-f)}{(d-a) (d-c)}=1$$ Plugging in $f=e^{-1},b=d^{-1},a=c^{-1}$ $$ LHS=\frac{{(c - e)(c - e^{-1})}}{{(c - d^{-1})(c - d)}} + \frac{{(d - e)(d - e^{-1})}}{{(d - c^{-1})(d - c)}} $$ After simplification, this is equal to $1$, we are done!
By symmetry, we can replace $c$ with $a$ in \eqref{3} to get another identity: $$\frac{(a-e)(a-f)}{(a-b)(a-d)}+\frac{(d-e) (d-f)}{(d-a) (d-c)}=1$$ In fact, the first term is invariant when replacing $c$ with $a$: $$\frac{(c-e) (c-f)}{(c-b) (c-d)}=\frac{(a-e)(a-f)}{(a-b)(a-d)}$$