Girls European Mathematical Olympiad

Since $\angle CDF=90^{\circ }-45^{\circ }=45^{\circ }$, the line $DF$ bisects $\angle CDA$, and so $F$ lies on the perpendicular bisector of segment $AE$, which meets $AB$ at $G$. Let $\angle ABC=2\beta$. Since $ADEF$ is cyclic, $\angle AFE=90^{\circ }$, and hence $\angle FAE=45^{\circ }$. Further, as $BF$ bisects $\angle ABC$, we have $\angle FAB=90^{\circ }-\beta$, and thus $\angle EAB=\angle AEG=45^{\circ }-\beta$ and $\angle AED=45^{\circ }+\beta$, so $\angle GED=2\beta$. This implies that right-angled triangles $△EDG$ and $△BDC$ are similar, and so we have $\frac{|GD|}{|CD|}=\frac{|DE|}{|DB|}$. Thus the right-angled triangle $△DEB$ and $△DGC$ are similar, whence $\angle GCD=\angle DBE=\beta$. But $\angle DFE=\angle DAE=45^{\circ }-\beta$, then $\angle GFD=45^{\circ }-\angle DFE=\beta$. Hence $GDCF$ is cyclic, so $\angle GFC=90^{\circ }$, whence $CF$ is perpendicular to the radius $FG$ of $\omega$. It follows that $CF$ is a tangent to $\omega$, as required.

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Alternative solution.

As $\angle ADF=45^{\circ }$ line $DF$ is an exterior bisector of $\angle CDB$. Since $BF$ bisects $\angle DBC$ line $CF$ is an exterior bisector of $\angle BCD$. Let $\angle ABC=2\beta$, so $\angle ECF=(\angle DBC+\angle CDB)/2=45^{\circ }+\beta$. Hence $\angle CFE=180^{\circ }-\angle ECF-\angle BCE-\angle EBC=180^{\circ }-(45^{\circ }+\beta +90^{\circ }-2\beta +\beta )=45^{\circ }$. It follows that $\angle FDC=\angle CFE$, then $CF$ is tangent to $\omega$.

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Alternative solution.

Note that $AE$ is a diameter of circumcircle of $△ABC$ since $\angle CDF=90^{\circ }$. From $\angle AEF=\angle ADF=45^{\circ }$ it follows that triangle $△AFE$ is right-angled and isosceles. Without loss of generality, let points $A,E$ and $F$ have coordinates $(-1,0),(1,0)$ and $(0,1)$ respectively. Points $F,E,B$ are collinear, hence $B$ have coordinates $(b,1-b)$ for some $b\ne -1$. Let point $C^{\prime }$ be intersection of line tangent to circumcircle of $△AFE$ at $F$ with line $ED$. Thus $C^{\prime }$ have coordinates $(c,1)$ and from $C^{\prime }\cdot E\perp AB$ we get $c=\frac{2b}{b+1}$. Now vector $\overrightarrow{B{C}^{\prime }}=(\frac{2b}{b+1}-b,b)=\frac{b}{b+1}(1-b,b+1)$, vector $\overrightarrow{BF}=(-b,b)=b(-1,1)$ and vector $\overrightarrow{BA}=-(b+1),-(1-b))$. Its clear that $(1-b,b+1)$ and $(-b+1),-(1-b))$ are symmetric with respect to $\overrightarrow{FE}=(-1,1)$, hence $BF$ bisects $\angle C^{\prime }BA$ and $C^{\prime }=C$ which completes the proof.

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Alternative solution.

Again $F$ lies on the perpendicular bisector of segment $AE$, so $△AFE$ is right-angled and isosceles. Let $M$ be an intersection of $BC$ and $AF$. Note that $△AMB$ is isosceles since $BF$ is a bisector and altitude in this triangle. Thus $BF$ is a symmetry line of $△AMB$. Then $\angle FDE=\angle FEA=\angle MEF=45^{\circ }$, $AF=FE=FM$ and $\angle DAE=\angle EMC$. Let us show that $EC=CM$. Indeed, $$\angle CEM=180^{\circ }-(\angle AED+\angle FEA+\angle MEF)=90^{\circ }-\angle AED=DAE=\angle EMC.$$ It follows that $FMCE$ is a kite, since $EF=FM$ and $MC=CE$. Hence $\angle EFC=\angle CFM=\angle EDF=45^{\circ }$, so $FC$ is tangent to $\omega$.

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Alternative solution.

Let the tangent to $\omega$ at $F$ intersect $CD$ at $C^{\prime }$. Let $\angle ABF=\angle FBC=\beta$. It follows that $\angle C^{\prime }FE=45^{\circ }$ since $C^{\prime }F$ is tangent. We have $$\frac{\sin \angle BDC}{\sin \angle CDF}\cdot \frac{\sin \angle DF{C}^{\prime }}{\sin \angle C^{\prime }FB}\cdot \frac{\sin \angle FBC}{\sin \angle CBD}=\frac{\sin 90^{\circ }}{\sin 45^{\circ }}\cdot \frac{\sin (90^{\circ }-\beta )}{\sin 45^{\circ }}\cdot \frac{\sin \beta }{\sin 2\beta }=\frac{2\sin \beta \cos \beta }{\sin 2\beta }=1.$$ So by trig Cheva on triangle $△BDF$, lines $F{C}^{\prime },DC$ and $BC$ are concurrent (at $C$), so $C=C^{\prime }$. Hence $CF$ is tangent to $\omega$.