Circumscribed quadrilateral

[ins-]

Let the quadrilateral $ABCD$ is circumscribed around a circle with center $I$. $E$ is the intersection point of $AB$ and $CD$. $F$ is the intersection point of $AD$ and $BC$. $J$ is the intersection point of the diagonals $AC$ and $BD$. Prove that $EJ^2+FI^2=EI^2+FJ^2$.

(This problem is inspired by Milen Naydenov's problem published in the Bulgarian math magazine "Mathematical forum")

[ins-]

It is too easy - don't loose your time.

[Dimoniada]

Interesting... I suppose, that $J$ was orthocenter $\triangle{EFI}$, but it's not (it is in inscribed quadrilateral).
Solution: let's $K,L,M,N$ - common tangency point of circle and sides $ABCD$ (look pic.) Then, obviously, lines $KL$, $MN$ pass through point $J$ and $EI^2 - FI^2 = EN^2 - FL^2$. Вy Stewart's Theorem for $\triangle{ENM}$ and $\triangle{FLK}$ we have $EJ^2 = -NJ{\cdot}JM + EN^2$ and $FJ^2 = -LJ{\cdot}JK + FL^2$, so $EJ^2+FI^2=EI^2+FJ^2$

[ins-]

I think $IJ$ is perpendicular to $EF$. But I rediscovered the problem statement starting from totally different initial position.

[Dimoniada]

Yes, $FE$ - polar of $J$ and previous mine solution is noob's one