Taiwan IMO Selection Camp

(1) Let $P, Q, R, S$ denote respectively the feet of the perpendiculars from $E$ to $DA, AB, BC, CD$. Note that $P, Q, A, E$ are concyclic, so $\angle QPE = \angle QAE$. Similarly, $\angle QRE = \angle QBE$.
Therefore, $\angle QPE + \angle QRE = \angle QAE + \angle QBE = 90^\circ$ (since $E$ lies on the circle with diameter $AB$.) Similarly, $\angle SPE = \angle SRE = 90^\circ$, hence $\angle QPS + \angle QRS = 90^\circ + 90^\circ = 180^\circ$, so $P, Q, R, S$ all lie on $\omega_E$. By symmetry, the four feet of perpendicular from $F$ to the four sides of $ABCD$ also all lie on $\omega_F$.

(2) Extend $AD$ and $BC$ to meet at $K$; without loss of generality, assume $A$ lies on segment $DK$. We show below that $\angle CKD$ is acute.
If not, then the circle with diameter $CD$ would contain $\triangle CKD$, and hence would contain the entire quadrilateral $ABCD$, so that $E, F$ could not lie inside the quadrilateral $ABCD$, a contradiction. Therefore, $\angle CKD$ is acute, and thus the line $EP$ must meet segment $BC$ at some point $P'$, and the line $ER$ must also meet segment $AD$ at some point $R'$.

(3) We next show that $P'$ and $R'$ also lie on $\omega_E$. The proof is as follows.
Note that $R, E, Q, B$ are concyclic, so by (1),
$$\angle QRK = \angle QRB = \angle QEB = 90^\circ - \angle QBE = \angle QAE = \angle QPE = \angle QPP',$$
which shows that $P'$ lies on the circle $\omega_E$. Similarly, $R'$ also lies on the circle $\omega_E$.

(4) Analogous to (1) through (3), let $M, N$ denote respectively the feet of the perpendiculars from $F$ to $AD$ and $BC$, and let $M' = FM \cap BC, N' = FN \cap AD$. By the same reasoning, we know that $M', N'$ both lie on $\omega_F$.

Figure 2

(5) Now, let $U$ be the intersection point of $NN'$ and $PP'$, and let $V$ be the intersection point of $MM'$ and $RR'$. Note that $N, P', N', P$ are concyclic, so $UN \times UN' = UP \times UP'$, and hence $U$ lies on the line joining the two intersection points of circles $\omega_E$ and $\omega_F$. Similarly, $V$ also lies on the line joining the two intersection points of circles $\omega_E$ and $\omega_F$.

(6) Finally, since $EUFV$ is a parallelogram, the midpoint of $EF$ lies on $UV$, and hence lies on the line joining the two intersection points of circles $\omega_E$ and $\omega_F$. Q.E.D.