It is obviously enough to prove that $C$ lies on symmedian of $\Delta ADB$ (because symmedian contains intersection of tangents), so we let $G$ be a midpoint of $DB$. Now it remains to prove that $\Delta ADG \sim \Delta ACB$ or $\frac{AD}{DG}=\frac{AC}{CB}$ $\iff$ $\frac{AD}{AC}=\frac{DG}{CB}$ $\iff$ $\frac{AD}{AC}=\frac{BG}{CB}$ $\iff$ $\Delta ADC \sim \Delta GBC$. But, by easy angle chase, we obtain that $\Delta ADC \sim \Delta EBC$, thus we need to prove $\Delta GBC \sim \Delta BEC$ or $\frac {GB}{BC} = \frac{BE}{CE}$ $\iff$ $\frac {\frac{1}{2}DB}{BC} = \frac{\frac{1}{2}AE}{CE}$ $\iff$ $\frac {DB}{BC} = \frac{AE}{CE}$ $\iff$ $\Delta AEC \sim \Delta DCB$. We will prove this. We have $\angle DBC =\angle EDF =\angle CEA$ and $\angle BDC= \angle CAE$ thus $\Delta AEC \sim \Delta DCB$. Q.E.D

Since $B$ is midpoint and $DF\parallel BA$ then $D(A,C;B,F)=-1 $ but $DF$ is tangent to $(ABCD)$ so $ACBD$ is harmonic hence $AC$ pass through $F$. R HAS PS: if the students knows harmonic topic it s very easy for the selection of the team.

Since $DF\parallel AB$ we get $\measuredangle CAD=\measuredangle DEA$ $\Longrightarrow$ $\triangle BCE\sim \triangle DCA$ $\Longrightarrow$ $\tfrac{AB}{BC}=\tfrac{EB}{BC}=\tfrac{AD}{CD}$ $\Longrightarrow$ $AB.CD=BC.AD$ $\Longrightarrow$ $ABCD$ is a quadrilateral armonic hence $A$, $C$ and $F$ are collinear.

PROF65 wrote: Since $B$ is midpoint and $DF\parallel BA$ then $D(A,C;B,F)=-1 $ but $DF$ is tangent to $(ABCD)$ so $ACBD$ is harmonic hence $AC$ pass through $F$. R HAS PS: if the students knows harmonic topic it s very easy for the selection of the team. Actually, only few students knew projective geometry in our country and as I recall only one contestant solved it with projective geometry

gobathegreat wrote: Let $ABCD$ be a quadrilateral inscribed in circle $k$. Lines $AB$ and $CD$ intersect at point $E$ such that $AB=BE$. Let $F$ be the intersection point of tangents on circle $k$ in points $B$ and $D$, respectively. If the lines $AB$ and $DF$ are parallel, prove that $A$, $C$ and $F$ are collinear. I have a solution with complex numbers. Let the circle $k$ be the unit circle. $\boxed{1}$ Since $DF$ is tangent to the circle $k$ and $AB \parallel DF,$ \[\angle CAD=\angle FDC=\angle DEA.\]It means that $\triangle EAD \sim \triangle ACD$ $\Longrightarrow$ $\angle BAD=\angle ACD=\angle ABD$ $\Longrightarrow$ $AD=BD.$ So, we get \[ (a-d)(\overline{a}-\overline{d})=(b-d)(\overline{b}-\overline{d}).\]$\Longrightarrow$ \[\frac{(a-d)^2}{a}=\frac{(b-d)^2}{b}.\]$\Longrightarrow$ \[b-a=\frac{d^2}{ab}(b-a)\]$\Longrightarrow$ \[ab=d^2.\]$\boxed{2}$ Since $E=AB\cap CD,$ \[e=\frac{ab(c+d)-cd(a+b)}{ab-cd}.\]Also, $B$ is the midpoint of $EA,$ so \[b=\frac{e+a}{2}=\frac{\frac{ab(c+d)-cd(a+b)}{ab-cd}+a}{2}\]$\Longrightarrow$ \[2(ab^2+cad)=abc+bcd+abd+a^2b\]$\Longrightarrow$ because $ab=d^2,$ \[2(d^2b+cad)=d^2c+bcd+abd+ad^2\]$\Longrightarrow$ \[2(bd+ca)=cd+bc+ab+ad.\] $\boxed{3}$ Now, we are trying to show that $A,C,F$ are collinear $\iff$ \[\frac{a-c}{\overline{a}-\overline{c}}=\frac{c-f}{\overline{c}-\overline{f}}.\]Because $F$ is the intersection of tangents at $B$ and $C,$ $f=\frac{2bd}{b+d}$ and $\overline{f}=\frac{2}{b+d}.$ So, \[\frac{a-c}{\overline{a}-\overline{c}}=\frac{c-f}{\overline{c}-\overline{f}}\]$\iff$ \[-ac=\frac{c-\frac{2bd}{b+d}}{\frac{1}{c}-\frac{2}{b+d}}\]$\iff$ \[c-\frac{2bd}{b+d}=-a+\frac{2ac}{b+d}\]$\iff$ \[bc+ab+cd+ad=2(ac+bd),\]which is equivalent to $\boxed{2}.$ So, the proof is completed.

Hello. Let $M$ be the midpoint of $AC$.It suffices to show that $BD$ is the symmedian of the triangle $\triangle{ABC}$. Since $B$ is the midpoint of $AE$ we have $\angle{ABM}=\angle{AED}\overset{DF\parallel AB}=\angle{FDE}=\angle{DBC}$,where the last equality is obtained from the alternate secant theorem,and we are done.

Hello can you explain the operation in the first box after getting that AD=BD ?

Can someone explain this result?

From tangent to angle, $\angle EDF=\angle DAC.$ But from $DF||BA$ we have $\angle EDF=\angle DFA,$ so $\angle DFA=\angle DAC$ and thus $\triangle ADE\sim\triangle CDA\sim \triangle CBE\implies \dfrac{AD}{DE}=\dfrac{CD}{DA}=\dfrac{CB}{BE}=\dfrac{CB}{BA}\implies ABCD$ is harmonic, and thus by a well known lemma $C$ lies on the $A-$symmedian of $\triangle ABD. \blacksquare$

Great solution jaobundao

Let $FC\cap k=M$ and $FC\cap BD=N$ So we have $(F,N;C,M)=-1$. $\Longrightarrow$ $(P_{\infty},B;E,DM\cap AE)=-1$ But we have $(P_{\infty},B;E,A)=-1$ So $A,F,C$ are collinear.

As $AE\|DF$ and $B$ is the midpoint of $AE$. Hence, $$-1=(A,E;B,P_{\infty})\overset{D}{=}(A,C;B,D)$$. Hence, $A-C-F$ are collinear.

We have that $D$ is the midpoint of arc $AB$, so $AD=BD$. Also, triangles $ABD$ and $DBF$ are similar, so: $\frac{BD}{AB}=\frac{FD}{AD}$. Another similar triangles are $ACE$ and $DBE$, so using that $AB=BE$ and what we previously said we obtain: $\frac{AC}{CE}=\frac{BD}{BE}=\frac{BD}{AB}=\frac{FD}{AD}$. $\angle ADF=180^\circ-\angle BAD=180^\circ-\angle ABD=180^\circ-\angle ACD=\angle ACE \implies$ triangles $ACE$ and $ADF$ are similar, so $\angle EAC=\angle CFD$, so $A,C,F$ are collinear.

Posting this one for storage Since we have that $DF \parallel AB$ and because $AB=BE$ we have that: $$-1=(AE;BP_{\infty}) \overset{D}{=} (AC;BD)$$ Thus this implies that $ABCD$ is a harmonic quad, thus we have that the points $A,C$ and $F$ are colinear.