Junior Mathematical Olympiad

Note that the triangle $ABC$ is isosceles right triangle. Let $CD \cap AB = \{H\}$. From
$$\angle AED = \angle ADB = 90^\circ \text{ and } \angle DAE = \angle DAB$$
follows that $\triangle ADE \sim \triangle ABD$. Since $ABCD$ is cyclic, it follows
$$\angle ADC = 180^\circ - \angle ABC = 135^\circ \text{ i.e. } \angle HDA = 45^\circ \text{. Hence,}$$
$$\angle DAB = \angle DAC + \angle CAB = \angle DBC + \angle CAB = 22^\circ 30' + 45^\circ = 67^\circ 30'$$
Now we have,
$$\angle HAD = 180^\circ - \angle DAB = 180^\circ - 67^\circ 30' = 112^\circ 30'$$
and
$$\angle AHD = 180^\circ - (\angle HAD + \angle HDA) = 180^\circ - 157^\circ 30' = 22^\circ 30'$$
i.e. $\triangle HDB$ is isosceles. Since $\triangle HDB$ is isosceles and $DE \perp AB$ it follows that $E$ is a midpoint of $HB$. Since $E$ is a midpoint of $HB$ and $F$ is a midpoint of $CB$, follows that $EF$ is a median line in $\triangle HBC$ i.e. $EF \parallel HC$ i.e. $EF \parallel CD$. Hence, the quadrilateral

$EFCD$ is trapezoid. Furthermore, $EF \parallel HC$, so we have $\angle FEB = \angle CHB = 22^\circ 30'$. From $DE \perp AB$ we have
$$\angle DEF = 90^\circ - \angle FEB = 90^\circ - 22^\circ 30' = 67^\circ 30'$$
Then,
$$\angle EFB = 180^\circ - (\angle FEB + \angle EBF) = 180^\circ - 67^\circ 30' = 112^\circ 30' \text{ i.e.}$$
$$\angle CFE = 180^\circ - \angle EFB = 67^\circ 30'$$
Finally, $EFCD$ is an isosceles trapezoid, hence the statement in the problem follows.