Solution:
We first prove the following lemma:
Lemma 1. Let $A B C D$ be a convex quadrilateral and let $A B \cap C D=E$ and $B C \cap D A=F$. Then the circumcircles of triangles $A B F, C D F, B C E$ and $D A E$ all pass through a common point $P$. This point lies on line $E F$ if and only if $A B C D$ is concyclic.
Proof. Let the circumcircles of $A B F$ and $B C F$ intersect at $P \neq B$. We have
$$\text{F P C = F P B+ B P C= B A D+ B E C= E A D+ A E D= =180 - A D E=180 - F D C}$$
which gives us $F, P, C$ and $D$ are concyclic. Similarly we have
$$\text{A P E = A P B+ B P E= A F B+ B C D= D F C+ F C D= =180 - F D C=180 - A D E}$$
which gives us $E, P, A$ and $D$ are concyclic. Since $\text{F P E= F P B+ E P B= B A D+}$ $\text{B C D}$ we get that $\text{F P E=180}$ if and only if $\text{B A D+ B C D=180}$ which completes the lemma.

We now divide the problem into cases:

Case 1: $A E P F$ and $B F E C$ are concyclic. Here we get that
$$\text{180 = A E P+ A F P=360 - C E B- B F C=360 -2 C E B}$$
and here we get that $\text{C E B= C F B=90}$, from here it follows that $P$ is the orthocenter of $\triangle A B C$ and that gives us $\text{A D B= A D C=90}$. Now the quadrilaterals $C E P D$ and $B D P F$ are concyclic because
$$\text{C E P= C D P= P D B= P F B=90 .}$$
Quadrilaterals $A C D F$ and $A B D E$ are concyclic because
$$\text{A E B= A D B= A D C= A F C=90}$$

Case 2: $A E P F$ and $C E P D$ are concyclic. Now by lemma 1 applied to the quadrilateral $A E P F$ we get that the circumcircles of $C E P, C A F, B P F$ and $B E A$ intersect at a point on $B C$. Since $D \in B C$ and $C E P D$ is concyclic we get that $D$ is the desired point and it follows that $B D P F, B A E D, C A F D$ are all concyclic and now we can finish same as Case 1 since $A E D B$ and $C E P D$ are concyclic.

Case 3: $A E P F$ and $A E D B$ are concyclic. We apply lemma 1 as in Case 2 on the quadrilateral $A E P F$. From the lemma we get that $B D P F, C E P D$ and $C A F D$ are concyclic and we finish off the same as in Case 1.

Case 4: $A C D F$ and $A B D E$ are concyclic. We apply lemma 1 on the quadrilateral $A E P F$ and get that the circumcircles of $A C F, E C P, P F B$ and $B A E$ intersect at one point. Since this point is $D$ (because $A C D F$ and $A B D E$ are concyclic) we get that $A E P F, C E P D$ and $B F P D$ are concyclic. We now finish off as in Case 1. These four cases prove the problem statement.

Remark. A more natural approach is to solve each of the four cases by simple angle chasing.