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 $△ADE\sim △ABD$. Since $ABCD$ is cyclic, it follows $$\angle ADC=180^{\circ }-\angle ABC=135^{\circ }\text{ i.e. }\angle HDA=45^{\circ }\text{. Hence,}$$ $$\angle DAB=\angle DAC+\angle CAB=\angle DBC+\angle CAB=22^{\circ }30^{\prime }+45^{\circ }=67^{\circ }30^{\prime }$$ Now we have, $$\angle HAD=180^{\circ }-\angle DAB=180^{\circ }-67^{\circ }30^{\prime }=112^{\circ }30^{\prime }$$ and $$\angle AHD=180^{\circ }-(\angle HAD+\angle HDA)=180^{\circ }-157^{\circ }30^{\prime }=22^{\circ }30^{\prime }$$ i.e. $△HDB$ is isosceles. Since $△HDB$ 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 $△HBC$ i.e. $EF\parallel HC$ i.e. $EF\parallel CD$. Hence, the quadrilateral

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