I am interested in a question by @guangzhou-2015 with no solutions. Equivalent form:
An ellipse with foci $E,F$ is tangent to four sides of quadrilateral $ABCD$, then $$\sqrt{EA \times EC \times FA \times FC}+\sqrt{EB \times ED \times FB \times FD}=\sqrt{AB \times BC \times CD \times DA}$$
By dimensional analysis, the quantity in this equation is square root of 4th power of length, so I guess it is a relation of area.
This equation is equivalent to
$$\sqrt{EA \times EC \times FA \times FC}+\sqrt{EB \times ED \times FB \times FD}=\text{constant}$$for all isogonal conjugate pairs $(E,F)$ wrt convex quadrilateral $ABCD$ and inside it.
We get the right hand side if we evaluate this expression at $(A,C)$ or $(B,D)$.
The points $P,Q$ should be restricted to inside the quadrilateral, otherwise the square roots need to change sign. In the "foci" form of this question, we implicitly assumed $P,Q$ are inside $ABCD$ by using an "ellipse" rather than an arbitrary conic.
Using $\rm(eq1)$ by symmetry $$ \frac{AE}{AB}=\frac{CF⋅DE}{BF⋅CD} $$ and $$ \frac{AF}{AD}=\frac{FB⋅CE}{BC⋅ED} $$ we get \begin{align} \sqrt{AE⋅AF⋅CE⋅CF\over AB⋅AD⋅CB⋅CD}&= \sqrt{\frac{CF⋅DE}{BF⋅CD}⋅\frac{FB⋅CE}{BC⋅ED}⋅\frac{CE⋅CF}{CB⋅CD}}\\ &=\frac{CE⋅CF}{CB⋅CD} \end{align} Similarly $$ \sqrt{BE⋅BF⋅DE⋅DF\over AB⋅AD⋅CB⋅CD}=\frac{DE⋅DF}{DA⋅DC} $$ Substituting into $\rm(eq2)$ we get $$ \sqrt{AE⋅AF⋅CE⋅CF\over AB⋅AD⋅CB⋅CD}+\sqrt{BE⋅BF⋅DE⋅DF\over AB⋅AD⋅CB⋅CD}=1 $$ Therefore $$ \sqrt{AE⋅AF⋅CE⋅CF}+\sqrt{BE⋅BF⋅DE⋅DF}=\sqrt{AB⋅AD⋅CB⋅CD} $$