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

$$\begin{aligned} \angle F P C & =\angle F P B+\angle B P C=\angle B A D+\angle B E C=\angle E A D+\angle A E D= \\ & =180^{\circ}-\angle A D E=180^{\circ}-\angle F D C \end{aligned}$$

which gives us $F, P, C$ and $D$ are concyclic. Similarly we have

$$\begin{aligned} \angle A P E & =\angle A P B+\angle B P E=\angle A F B+\angle B C D=\angle D F C+\angle F C D= \\ & =180^{\circ}-\angle F D C=180^{\circ}-\angle A D E \end{aligned}$$

which gives us $E, P, A$ and $D$ are concyclic. Since $\angle F P E=\angle F P B+\angle E P B=\angle B A D+$ $\angle B C D$ we get that $\angle F P E=180^{\circ}$ if and only if $\angle B A D+\angle B C D=180^{\circ}$ 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

$$180^{\circ}=\angle A E P+\angle A F P=360^{\circ}-\angle C E B-\angle B F C=360^{\circ}-2 \angle C E B$$

and here we get that $\angle C E B=\angle C F B=90^{\circ}$, from here it follows that $P$ is the orthocenter of $\triangle A B C$ and that gives us $\angle A D B=\angle A D C=90^{\circ}$. Now the quadrilaterals $C E P D$ and $B D P F$ are concyclic because

$$\angle C E P=\angle C D P=\angle P D B=\angle P F B=90^{\circ} .$$

Quadrilaterals $A C D F$ and $A B D E$ are concyclic because

$$\angle A E B=\angle A D B=\angle A D C=\angle A F C=90^{\circ}$$

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.

## Number Theory