Discussed before at http://www.artofproblemsolving.com/community/c6h411766.

Let the feet of the perpendiculars from $A,B,C,D$ to $CD,$ $DE,$ $EA,$ $AB$ is $A',B',C',D'$ respectively and they concur at $P.$ And let $EP$ meet the line $BC$ at $E'$. We are going to show that $E' \in (BPD')$. We can easly see that $A,D',P,C'$ ; $E,B',P,C'$ and $D,A',P,B'$ are cyclic. So if we write powers of $A,B,C,D$ wrt these circle we get \[ AP\cdot PA'=BP\cdot PB'=CP\cdot PC'=DP\cdot PD'=PR^2 \]Let $\gamma$ be the $PR$-radius circle with center $P$. Inversion wrt $\gamma$ swaps $A,B,C,D$ with $A',B',C',D'$ respectively. So the line $BC$ maps to $(B'PC')$. $E\in (B'PC') \implies EP\cdot E'P=PR^2 \implies E',B,D',P $ is cyclic.

Here is a short sketch of another solution with using orthologic triangles. Just observe $\triangle (AE,ED,CD)$ and $\triangle BAC$ are orthologic triangles with ortgology center $P.$