My solution: Let $ S \in BF $ be the incenter of $ \triangle BCD $ . From $ \angle FDA=45^{\circ} $ we get $ DF $ is the external bisector of $ \angle BDC $ , so $ F $ is the B-excenter of $ \triangle BCD \Longrightarrow C, D, F, S $ are concyclic , hence $ \angle DAE=\angle DFS=\angle DCS \Longrightarrow AE \perp CS \Longrightarrow AE \parallel CF $ . ... $ (\star) $ Since $ \angle FEA=\angle FDA=45^{\circ}, \angle AFE=90^{\circ} $ , so we get the center of $ \omega $ is the midpoint $ T $ of $ AE $ and $ TF \perp AE $ , hence combine with $ (\star) $ we get $ CF $ is tangent to $ \omega $ at $ F $ . Q.E.D ____________________________________________________________ EDIT : simpler solution ( thanks for navredras pointed me out ) After we get $ C, D, F, S $ are concyclic we can finish the proof as following : From $ \angle EFC=\angle SDC=45^{\circ}=\angle EDF $ we get $ CF $ is tangent to $ \omega $ at $ F $ . Done

We easily get that $DF$ is the angle bisector of $\angle EDA$ so $F$ is equidistant from $AB$ and $CD$. $F$ also belongs to angle bisector of $\angle ABC$ so it is also equidistant from $AB$ and $BC$. From this we have that $F$ is the $B$-excenter of triangle $BCD$. From this we get $\angle BFC= 180 - \angle CBF + \angle BCE + \angle ECF = 45 $ so we are finished.

Let $\angle ABC=x$ and $\angle EAC=y.$ Since angle bisectors of $\angle ADC$ and $\angle ABC$ intersect at point $F$, obviously $F $ is the excenter of the triangle $BDC$. Hence $\angle DCF 45^0+x/2$(because $\angle BCD=90^0-x $. By easy angle chasing we obtain that $\angle ECA=45^0+x/2-y$. Which leads to $\angle ACF=y$. And $ AE $ parallel to $ CF $. So $\angle AEF=\angle EFC=45^0=\angle EDF$. Which clearly shows that $ CF $ tangent to $ w $

my solution = since $\angle ADF=90-\angle CDB$ thus $DF$ is external angle bisector of $\angle CDB$ thus $F$ is $B$-excentre of triangle $CBD$. hence $\angle DCF=90-\frac{\angle BCD}{2}=135-\frac{\angle B}{2}$ so,$\angle BFC=45=\angle EDF=\angle EAF$ as $A,D,E,F$ are concyclic. and hence $CF$ is tangent to $\omega$ so we are done

Suppose $AF$ intersects $BC$ at $A'$. $BA=BA'$ and $BF\perp AA'$, so $\angle EA'F=\angle EAF=\angle ADF=45$, i.e. $A'FE$ is a 45-45-90 triangle. Angle chasing yields $\angle CAE=\angle CEA$, implying that $CF$ is the perpendicular bisector of $A'E$. Thus $\angle CFE=45=\angle FDE$, i.e. $CF$ is tangent to $\omega$.

Let $CM$ be the angle bisector of $\angle DCB$ with $M \in AB$. Furthermore let $I$ be the incenter of $\triangle DBC$ and $S = AE \cap CM$. We know that $\angle FEA = 45^{\circ}$ then $\angle AEB = 135^{\circ}$ which means $$\dfrac{\angle ABC}{2} = \angle ABE = 180^{\circ} - 135^{\circ} - \angle EAB = 45^{\circ} - \angle EAB.$$ Then $\angle ABC = 90^{\circ} - 2\angle EAB$ implying $\angle DCB = 2\angle EAB$. Now since $CM$ is the angle bisector of $\angle DCB$ we know $\angle DCM = \angle EAB$. Therefore $\triangle ADE \sim \triangle CDM$ so we can rotate $\triangle ADE$ 90 degrees clockwise around $D$ and then do a homothety to get $\triangle CDM$. This means that any two corresponding lines in the two triangles are perpendicular, in particular $AE \perp CM$. Combining this with $CD \perp AB$ we easily get quadrilateral $ADSC$ is cyclic. This gives $CE \cdot ED = AE \cdot ES$. Furthermore since we know $\angle ESI = EFA = 90^{\circ}$ quadrilateral $FAIS$ is cyclic as well. This gives $FE \cdot EI = AE \cdot ES$. Therefore $CE \cdot ED = FE \cdot EI$ and so quadrilateral $FDIC$ is cyclic. This gives $\angle EFC = \angle IFC = \angle IDC = 45^{\circ}$ because $I$ is the incenter of $\triangle DCB$. Therefore $\angle EFC = \angle FAE = 45^{\circ}$ which gives $CF$ is tangent to the circumcircle of $\triangle AED$ as desired.

Construct $X$ on $BC$ such that the circle with radius $DE$ and center $E$ is the incircle of $\triangle ABX$. Since $\angle BEA = 135^\circ$, we have that \[\angle BXA = 180^\circ - 2(\angle EBA + \angle EAB) = 180^\circ - 2 \cdot 45^\circ = 90^\circ.\]So $BXFA$ is cyclic, so $\angle FAX = \angle FBX = \angle FBA = \angle FXA \implies FX = FA = FE$. Thus $\angle XCE = 90^\circ - \angle CBD = \angle XAB = \angle XFB$, so $CXEF$ is cyclic. So we have $\angle CFE = \angle BXE = \tfrac{\angle AXB}{2} = 45^\circ$, so $\angle CFE = \angle FAE \implies CF$ is tangent to $\omega$, as desired.

Hmm, AoPS does not appear to support opacity [asy][asy] import olympiad; import cse5; size(8cm); defaultpen(fontsize(9pt)); pair A = dir(0); pair B = dir(180); pair F = dir(65); pair D = extension(A, B, F, dir(205)); pair K = F*F; pair C = extension(D, D+dir(90), B, K); pair E = extension(B, F, C, D); filldraw(unitcircle, palecyan+opacity(0.2), heavyblue); filldraw(circumcircle(A, D, E), palered+opacity(0.2), red+dotted); filldraw(CP(E,D), pink+opacity(0.2), magenta); draw(arc(F,abs(E-F),170,310), orange+dashed); draw(K--A--B--C--D, deepcyan); draw(C--A, deepcyan+dashed); draw(B--F--A, deepgreen); draw(F--D, deepgreen); draw(C--F, mediumred); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$F$", F, dir(F)); dot("$D$", D, dir(225)); dot("$K$", K, dir(K)); dot("$C$", C, dir(C)); dot("$E$", E, dir(165)); [/asy][/asy] Let $BC$ meet the circle with diameter $\overline{AB}$ at $K$ By the conditions of the problem, we have $FK = FE = FA$. Thus $E$ is the incenter of $\triangle KBA$. By angle chasing, we can now show that $\angle KFE = 90^{\circ} - \frac12 \angle KEF = \angle BCD$, so $KCFE$ is cyclic and thus $\angle CKE = 135^{\circ} \implies \angle CFE = 45^{\circ}$ as needed. One can also realize $CKEF$ is cyclic by noting $BE \cdot BF = BD \cdot BA = BK \cdot BC$. $\blacksquare$. Alternatively, one can also finish using barycentric coordinates on $\triangle ABK$. Letting $a=BK$, $b=AK$, $c=AB$ we have $E = (a:b:c)$, $D = (s-b:s-a:0)$, and \[ F = (a(a+c) : -b^2 : c(a+c)) = (a(a+c) : a^2-c^2 : c(a+c)) = (a : a-c : c). \] But I guess I'll spare this problem...

Call the incenter of $\triangle BDC$ as $G$. Angle chasing makes us want $\angle FCG=90^\circ$, when we would be done from there. We have $BG/GE=BC/CE=BD/DE=BF/FA=BF/FE$ since $\overrightarrow {DF}$ bisects $\angle CDA$. Therefore $(B,E;G,F)$ is a harmonic bundle and from there it follows (provable with LoS) that $\angle FCG$ is right.

Produce $AF $ to meet $BC $ at $ Q $ $ \triangle ABF \cong \triangle QFB $ and $ \angle DCB = 90 - \angle B $ We can see that $F $ is the excenter of $ \triangle BDC $ opposite to side $ DC $ $ \Rightarrow \angle FCE = 45 + \angle B/2 $ $ \angle BQF = 90- \angle B/2 $ Now $ \angle CFQ+\angle CQF = \angle FCB =\angle FCE +\angle ECB $ $\Rightarrow \angle CFQ = 45 = \angle CFE $, so by alternate segment we get that $ C $ is tangent to $ \omega $

It's easy because we have $ CF || EA $ and $ AF=FE $.

my trigonometric solution

Since $ \angle ADF= \angle EDF=45^\circ$, we have $FA$=$FE$. Secondly. let's consider $A'$ the reflection of $A$ according to $BF$. Then we also have $FA$=$FA'$, so $FE$=$FA'$. By sines' law in $\triangle EFC$ and $\triangle A'FC$, we get $\frac{FC}{cos x}$=$\frac{FE}{sin \angle FCE}$ and $\frac{FC}{cos x}$=$\frac{FA'}{sin \angle FCA'}$, where $x=\frac{B}{2}$. From these two equations we get $\angle FCE$=$\angle FCA'$. And also $\angle FEC$=$\angle FA'C$, it means $\angle EFC$=$\angle A'FC$=$45^\circ$. So $\angle FDE$=$\angle EFC$, which exactly means that $CF$ is tangent to $\omega$. Note: If $sin \angle FCE$=$sin \angle FCA'$, then $\angle FCE$=$\angle FCA'$ or their sum is equal to $180^\circ$. But if their sum is equal to $180^\circ$, then $B$=$D$ or $\angle ABC$=$90^\circ$ and it is noted that the triangle is acute-angled. So, it is impossible and $\angle FCE$=$\angle FCA'$.

I have also solution Let $M=DF\cap AE$ and $AE\cap BC=T$ $BF\cap AC=K$ By using angle-chasing it suffices to prove that $FC || ME$ $\iff$ $\frac{FE}{EB}=\frac{CT}{TB}$ Here let $\measuredangle FBA=\alpha$ So we ll need to prove that $\cos (2\alpha)\cdot sin^2 (45^{\circ})=sin (45^{\circ}+\alpha)\cdot sin(45^{\circ}-\alpha)$ which is obvious.

Note that $F$ is the $B$-excenter of $\triangle CBD.$ It follows that $$\angle FED-\angle FDE=\left(90^{\circ}+\frac{\angle B}{2}\right)- 45^{\circ}=\left(45^{\circ}+\frac{\angle B}{2}\right)=\angle DCF,$$so $\overline{FC}$ is tangent to circle $\omega,$ as desired.

How can we guess about drawing the diagram which satisfies the condition. I have to draw so many times to get the right one. At first CF is almost the diameter of the circle!

Did you use ruler and compass? If you used, your diagram should be accurate.

Sorry for revive, but I think this is somewhat interesting:

I don’t think my solution is here so here it is

Let $P=AF\cap DE$, then $E$ is orthocenter of $APB$. Angle chasing with the orthocenter gives $\angle CPB=\angle EAB$, and $\angle EBP=\angle EAF$, implying $PCB$ isosceles with $CP=CB$. Also, $PFB$ is isosceles, with $PF=PB$, implying $C$, and $F$ are on the perpendicular bisector of $PB$, therefore, $FC$ is the angle bisector of $\angle BFP$. And since $\angle BFP=90^{\circ}$, then $CFE=45^{\circ}$, proving the tangent with inscribed angle $\angle FED=45^{\circ}$.

Yay I think I found a new solution, although it seems considerably more complicated than previous solutions. Firstly, notice $AE$ is the diameter of $\omega$, thus $\angle BFA =90^{\circ}$. Now let's angle chase in terms of $\angle ABC =\beta$. As $\angle ADF= 45^{\circ} => \angle FDE= 45^{\circ}$. As $\angle EBC= \frac{\beta}2$ and $\angle ECB= 90^{\circ}- \beta => \angle DEF= 90^{\circ}+\frac{\beta}2=> \angle EFD= 45^{\circ} -\frac{\beta}2$. Using $ADEF$ is cyclic and right angles, $\angle EAF = \angle EDF=45^{\circ}$ and $\angle FEA = \angle FDA = 45^{\circ}$, thus $\angle EAD =45^{\circ} - \frac{\beta}2 => \angle BAF = 90^{\circ} - \frac{\beta}2$. Here I struggled a bit till I got the idea of extending $AF$ to meet $BC$ at $X$- this almost immediately finishes off the problem. Noticing $\angle BXF= 180^{\circ} -(90^{\circ} + \frac{\beta}2) =90^{\circ}-\frac{\beta}2$, which turns out to equal $\angle BAX$, thus $AB=BX$- as $BF$ is perpendicular to $AX$, we get $AF=FX$, and as we already know $AF=FE => AF=FE=FX$. As $EF=FX$, and $\angle EFX=90^{\circ}=> \angle EXF = \angle FEX=45^{\circ}=\angle EAF => EA=AX$. As $\angle FEC=90^{\circ} -\frac{\beta}2 = \angle FXE => \angle CEX = 90^{\circ} - \frac{\beta}2 -45^{\circ}=45^{\circ} - \frac{\beta}2 = \angle CXE => CE=CX$. Thus by SSS $FEC \cong FXC => \angle EFC = \frac{\angle EFX}2=45^{\circ}$. Thus we're done by alternate tangent theorem as $\angle EFC = \angle EAC =45^{\circ}$. Nice problem, took some time even though it was only angle chasing.

Notation: Let $\angle B=2x$. One line proof: We have $\angle DCB=90-2x$ and thus, $\angle BEC= 90+x$, but, $\angle FCD = 45+x\implies \angle EFC = \angle FDE =45^\circ$ and we are done.

I believe this solution is new. Let $T$ be the point such that $E$ is the orthocenter of $\triangle TBC$, and define $X=BC\cap AT$. Let $\angle ABF=\angle FBX=\alpha$. Note that since $\angle BFA=90$, $\angle TAD=90-\alpha$ so $\angle FTE=\alpha$. Then $TXEB$ is cyclic. Now, as $\angle FEA=45$, $\angle TEY=45+\alpha$ so $\angle DTB=45-\alpha$. This implies that $\angle DBT=45+\alpha$ so that $\angle CBT=45-\alpha$. Thus $XE||TB$, which means $TXEB$ is in fact an isosceles trapezoid. Then by symmetry, $\triangle FXC\cong \triangle FEC$, which means $\angle XFC=\angle CFE=45$. Thus $CF$ is tangent to $(ADEF)$ by the alternate segment theorem.

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%207.pdf p. 91... Sincerely Jean-Louis

My Solution Step 1:$BF=AB\cdot \cos \alpha $ Proof: $ \angle AFB=\angle BDC=90 $ $ \angle BAF=90-\angle ABF=90-\alpha $ $ \frac{AB}{\sin 90}=\frac{BF}{\sin(90-\alpha)}\implies BF=AB\cdot \cos \alpha $ Step 2:$ CD=BC\cdot \sin 2\alpha $ Proof: $ \angle ABC=2\alpha $ $ \frac{BC}{\sin 90}=\frac{CD}{\sin 2\alpha}\implies CD=BC\cdot \sin 2\alpha $ Step 3:$BE=\sqrt{2}AB\cdot \sin(45-\alpha) $ Proof: $ \angle AEB=180-\angle AEF=180-\angle ADF=135 $ $ \angle BAE=\angle AEF-\angle ABE=45-\alpha $ $ \frac{AB}{\sin 135}=\frac{BE}{\sin (45-\alpha)}\implies BE=\sqrt{2}AB\cdot \sin(45-\alpha) $ Step 4:$ CE=\frac{\sqrt{2}\cdot AB\cdot \sin(45-\alpha)\cdot \sin \alpha}{\cos 2\alpha} $ Proof: $ \angle EBC=\alpha $ $ \angle DCB=90-2\alpha $ $ \frac{CE}{\sin \alpha}=\frac{BE}{\sin (90-2\alpha)}=\frac{\sqrt{2}AB\cdot \sin(45-\alpha)}{\sin (90-2\alpha)}\implies CE=\frac{\sqrt{2}\cdot AB\cdot \sin(45-\alpha)\cdot \sin \alpha}{\cos 2\alpha} $ Step 5:$ CF^2=AB^2\cdot \cos^2\alpha+BC^2-2AB\cdot BC\cdot \cos^2\alpha $ Proof: $ CF^2=BF^2+BC^2-2BF\cdot BC\cdot \cos \alpha=AB^2\cdot \cos^2\alpha+BC^2-2AB\cdot BC\cdot \cos^2\alpha $ Step 6:$AB=\frac{BC\cdot \cos 2\alpha}{\sqrt{2}\cdot \sin(45-\alpha)\cdot \cos \alpha} $ Proof: $ \angle BCD=90-2\alpha $ $ \angle BEC=90+\alpha $ $ \frac{BC}{\sin(90+\alpha)}=\frac{BE}{\sin(90-2\alpha)}=\frac{\sqrt{2}AB\cdot \sin(45-\alpha)}{\sin(90-2\alpha)}\implies AB=\frac{BC\cdot \cos 2\alpha}{\sqrt{2}\cdot \sin(45-\alpha)\cdot \cos \alpha} $ Step 7:$CF=\sqrt{2}BC\cdot \sin \alpha $ Proof: $AB=\frac{BC\cdot \cos 2\alpha}{\sqrt{2}\cdot \sin(45-\alpha)\cdot \cos \alpha} $ $ CF^2=AB^2\cdot \cos^2\alpha+BC^2-2AB\cdot BC\cdot \cos^2\alpha=2(BC\cdot \sin\alpha)^2\implies CF=\sqrt{2}BC\cdot \sin \alpha $ Step 8:$CE\cdot CD=2(BC\cdot \sin \alpha)^2 $ Proof: $AB=\frac{BC\cdot \cos 2\alpha}{\sqrt{2}\cdot \sin(45-\alpha)\cdot \cos \alpha} $ $ CE\cdot CD=\frac{\sqrt{2}\cdot AB\cdot \sin(45-\alpha)\cdot \sin \alpha}{\cos 2\alpha}\cdot BC\cdot \sin 2\alpha=2(BC\cdot \sin \alpha)^2 $ Finish: $CF^2=2(BC\cdot \sin \alpha)^2=CE\cdot CD $ $ CF^2=CE\cdot CD\implies CF$ is tangent to $\omega.$

Simple when you see it, a bit hard to find. Clearly, $F$ lies of the angle bisectors of $\angle ADC$ and $\angle ABC$. Thus, $F$ is an excenter of $\triangle CDB$. If we let $I$ be the incenter of $\triangle CDB$, by the incenter excenter lemma we have $\angle FCI= 90^{\circ}$. Notice that $\angle EIC = \angle IBC + \angle ICB= \dfrac{1}{2} \left ( \angle DBC + \angle BCD\right ) = \dfrac{1}{2} \left (90^{\circ} \right ) = 45^{\circ}$. So, $\angle CFI =45^{\circ}$. Clearly, $\angle FDC = 45^{\circ}$ which implies the result and concludes the proof. $\blacksquare$

Notice $F$ is the $B$-excenter of $\triangle BCD.$ Hence, \begin{align*}\angle ECF&=90-\tfrac{1}{2}\angle BCD\\&=45+\tfrac{1}{2}\angle DBC\\&=180-45-(90-\angle DBE)\\&=180-\angle FAD-\angle ADF\\&=\angle DFA\\&=\angle DEA\end{align*}and $\overline{CF}\parallel\overline{AE}.$ We see $\angle CFA=\angle AEF=\angle FDE.$ $\square$

oh god, took me so much time to realise that $F$ is the excenter.

is this even right Delete point $A$ because it is irrelevant. We wish to show that if $\triangle BDC$ is a right triangle with a right angle at $D$, and the B-bisector hits $CD$ at $E$, and has B-excenter $F$, then $CF$ is tangent to $(DEF).$ Note that $\angle FDC=45$. Let $\angle DBE=\theta$. Then, $\angle DEB=\angle FEC=90-\theta.$ Also, $\angle DCB=90-2\theta$, so $\angle DCF=45+\theta$, so $\angle EFC=45$. Since $$\angle FDC=\angle CFE=45,$$we are done.

My solution is also simmilar to other solutions $\angle CDA=90$ $\angle CDA=\angle CDF+\angle ADF$ $90=45+\angle CDF$ $\angle CDF=45$ $\angle ADF=\angle CDF$ $\implies$ $DF$ is angle bisector of $\angle CDA$ $...(1)$ $BE$ is the angle bisector of $\angle ABC$ $\implies$ $BE$ is also the angle bisector of $\angle CBD$ $...(2)$ $BE$ $\cap$ $DF$=$\{ F \}$ $\implies$ $F$ is the $B$-excenter of the $\triangle BDC$ because of $(1)$ and $(2)$ Since $F$ is the $B$-excenter we get that: $CF$ is the angle bisector of the external angle of $\angle BCD$ $\implies$ $\angle FCE$= $\dfrac{1}{2}(180- \angle BCD)$ $...(3)$ From the $\triangle BDC$ we get: $\angle BDC +\angle DBC +\angle BCD=180$ $90+\beta+\angle BCD=180$ $\angle BCD=90-\beta$ $...(4)$ Combining $(3)$ with $(4)$ we have: $\angle FCE$= $\dfrac{1}{2}(180- \angle BCD)=\dfrac{1}{2}(90+\beta)=45+\dfrac{\beta}{2}$ $\angle FCE=45+\dfrac{\beta}{2}$ By $\square ADEF-cyclic$ we have: $\angle FDE=\angle FAE=45=\angle FDA=\angle FEA$ $\implies$ $\angle FDE=\angle FDA=\angle FAE=\angle FEA=45$ Let $\angle DAE=x$ now from the $\triangle DAE$ we get: $\angle ADE+\angle DAE+\angle AED=180$ $90+x+\angle AED=180$ $\angle AED=90-x$ $\angle CED=180$ $\angle CED=\angle AED+\angle AEF+ \angle FEC$ $180=90-x+45+\angle FEC$ $45+x=\angle FEC$ $\angle FEC=45+x$ $\angle FEC=\angle BED$ $\angle BED=45+x$ $...(5)$ From $\triangle BDE$ $\angle BDE+ \angle EBD+ \angle BED=180$ $90+\dfrac{\beta}{2}+45+x=180$ $x=45-\dfrac{\beta}{2}$ Combining with $(5)$ we get: $\angle BED=90-\dfrac{\beta}{2}$ $\angle FEC=\angle BED=90-\dfrac{\beta}{2}$ $\angle FEC=90-\dfrac{\beta}{2}$ Now from the $\triangle CEF$ $\angle FEC+ \angle FCE+ \angle CFE=180$ $90-\dfrac{\beta}{2}+45+\dfrac{\beta}{2}+ \angle CFE=180$ $\angle CFE=45$ $\implies$ $\angle CFE=\angle FAE$ $<=>$ $CF$ is tangent to $w$

Let $\theta=\angle DBE$. Clearly $\angle AFE=90^{\circ}$, so $\angle EAF=\angle AEF=45^{\circ}$, so $\angle DFE=\angle DAE=45^{\circ}-\theta$. Now note that $\angle ADF=\angle CDF=45^{\circ}$ and $\angle DBF=\angle CBF$, so $F$ is the $B$-excenter of triangle $BCD$, so $\angle DFC=90^{\circ}-\theta$, so $\angle EFC=45^{\circ}=\angle EAF$ and we win. $\blacksquare$

Since $BE$ is angle bisector of $\angle{DBC}$ in $\triangle{BDC}$ we have incenter of $\triangle{BDC}$ $I \in BE$. Also since $\angle{ADF}=45^{\circ}$ we have $DF$ to be external angle bisector of $\angle{BDC}$ which gives $F$ to be $B-$ excenter of $\triangle{BCD}$. Now it is well known that $IDFC$ is cyclic , hence we have $\angle{IDC}=\angle{IFC}=\angle{FDE}=45^{\circ}$. Hence from the tangent secant theorem, we have $CF$ to be tangent to $\omega$, done. $\blacksquare$

Since $F$ lies on the angle bisectors of $\angle CBD$ and $\angle CDA$, it is the $B$-excenter in $\triangle BCD$. We have \[\angle CFB = 180^{\circ} - \frac{\angle B}{2} - \frac{180 - \angle C}{2} = 45^{\circ},\]and since $\angle FDE = 45^{\circ}$ too, the problem follows.