Note that it is enough to show that $\triangle{ABC} \sim \triangle{CFA}$. Let $L'$ be the isogonal conjugate of $L$ wrt $\triangle{DAC}$. Performing a reflection over the midpoint of segment $AC$ we see that $L'$ is the isotomic conjugate of $K$ wrt $\triangle{ABC}$, thus by Steiner's Theorem we have: $$\left(\frac{AB}{BC}\right)^2=\left(\frac{DC}{AC}\right)^2=\frac{CL\cdot CL'}{AL \cdot AL'}=\frac{CL\cdot AK}{AL \cdot CK} = (L,K;C,A) \stackrel{E}{=} (F,B;C,A)=\frac{FC\cdot AB}{FA \cdot BC}\implies \frac{FC}{FA}=\frac{AB}{BC} \implies \triangle{ABC} \sim \triangle{CFA} \ \ \ \blacksquare $$

Just intersect $CE $ and $AD $.And see two cyclic quadrilaterals. Is it really simple?I think it is harder than IMO 2017 P4.Quite hard for p1. (It took me one hour and half of hour)

Just note that $F$ and $D$ are symmetric about $AC$?

DVDthe1st wrote: Just note that $F$ and $D$ are symmetric about $AC$? Yeah.Official solution takes the reflection of D wrt AC. Then shows that it is just F.

Let $F'$ be the reflection of $D$ across $AC$. $AC$ passes through the midpoints of $DB$ and $DF’$. So, $AC \parallel BF'$. Also, $AB = CD = CF'$. So, $ABF'C$ is an isosceles trapezoid i.e. $ABF'C$ is cyclic. Now, $\angle CF'L = \angle CDL = \angle CBK = \angle CBE = \angle CF'E$. As $L$ aand $E$ lies on the same side of $F'$, so, $F', L, E$ are collinear. So, $F' = F$ and we got the desired result

Let $EV \parallel BC$ with $V$ on $\omega$. Reflect $L$ across mid-point of $AC$ to get $J$; then $BK, BJ$ are isogonal in angle $ABC$. Particularly, we see $B, J, V$ are collinear. Note that lines $BV$ and $EL$ are symmetric about the perpendicular bisector of $AC$. Consequently, $BFVE$ is an isosceles trapezoid, so $BF \parallel EV \parallel AC$. Remark. Motivated by "removing the floating $D$ from the picture". Also too easy for #1 in my opinion.

Let $AD\cap CE=S.$ and $\angle ABK=x,\angle CBK=y,\angle CAD=z,\angle CAB=t.$ We have $\angle ADL=\angle ABE=\angle ACS=x,$ then $DSCL$ is cyclic.Then $AL\cdot AC=AD\cdot AS,(1)$ since PoP. Also we can find easily ( with angle ) $\triangle LDC\sim \triangle CBE,(2)$ and $\triangle ABE\sim \triangle LDA.(3)$ Then $$AL\cdot AC=^{(1)} AD\cdot AS=BC\cdot AS=^{(2)} \frac{LD\cdot CE}{LC} \cdot AS=^{(3)} \frac{CE}{LC} \cdot AS\cdot \frac{AB\cdot LA}{AE}$$$\implies \frac{AC}{CB}\cdot \frac{AE}{AS}\cdot \frac{LC}{CE}=1\implies Sin(z+x)\cdot Sin(t)=Sin(x+t)\cdot Sin(z)\implies Cos(z+x-t)=Cos(x+t-z)\implies z=t.$ This means $t=\angle FAC=\angle FBC=z=\angle CAD=\angle ACB\to BF\parallel AC.$

Here's my synthetic solution by angle chasing: Let $BK$ intersects $\omega$ at $E$. Notice that it suffices to prove that $\measuredangle AEF = \measuredangle BEC$. Claim 01. $\triangle ALD \sim \triangle EAB$. Proof. Notice that \[ \measuredangle DAL \equiv \measuredangle DAC = \measuredangle BCA = \measuredangle BEA \]Notice that \[ \measuredangle LDA = \measuredangle ABK \equiv \measuredangle ABE \] Claim 02. $\triangle ALE \sim \triangle DLC$ Proof. Notice that by Claim 01, we have \[ \frac{AL}{AE} = \frac{DL}{AB} = \frac{DL}{DC} \]Using this and $\measuredangle EAL \equiv \measuredangle EAC = \measuredangle EBC \equiv \measuredangle KBC = \measuredangle CDL$, we have what we want. Now, to finish, notice that \[ \measuredangle AEF \equiv \measuredangle AEL = \measuredangle DCL \equiv \measuredangle DCA = \measuredangle BAC = \measuredangle BEC \]

a nice problem but easy let $F'$ be the reflection of $D$ on $AC$ claim(1):$BE'||AC$ proof: since $F'C=CD=AB$ and $\angle ACF'=\angle ACD=\angle CAB$ so $ABF'C$ is an isoscele trapezoid $\blacksquare$ claim(2):$F'=F$ proof: $\angle AF'C = \angle ADC =\angle ABC$ so $F' \in (ABC)$ $\angle AF'L=\angle ADL=\angle ABK=\angle ABE$ so $F',L,E$ are collinear $\blacksquare$

Let $F'$ denote the point on $\omega$ such that $BF' \parallel AC$. A solution purely with complex numbers is not difficult. Here, we just use it to prove the basic fact that $F'$ is the reflection of $D$ over $AC$. Note that $d = a + c - b$ and $f' = \frac{ac}{b}$. The reflection of $f'$ over $AC$ is given by \[ a + c -ac \overline{f} = a + c - b = d. \] Then the angle chase is immediate. \[ \angle AF'L = \angle ADL - \angle KBA = \angle EBA = \angle EFA \]so $F' = F$.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -30.522181898533855, xmax = 25.099726037859206, ymin = -5.046513103909196, ymax = 28.21045012698899; /* image dimensions */ /* draw figures */ draw((-11,1)--(-3,12), linewidth(0.4) + red); draw((-3,12)--(6,12), linewidth(0.4) + red); draw((6,12)--(-2,1), linewidth(0.4) + red); draw((-2,1)--(-11,1), linewidth(0.4) + red); draw((-11,1)--(6,12), linewidth(0.4) + red); draw((-6.639215718736752,3.8216839466997485)--(-2,1), linewidth(0.4) + red); draw((-3,12)--(-3.5077820305878022,5.847905744913775), linewidth(0.4) + red); draw(circle((-6.5,12.681818181818182), 12.51858123083674), linewidth(0.4) + linetype("4 4") + blue); draw((-18.942629407869116,11.30491969191832)--(-6.639215718736752,3.8216839466997485), linewidth(0.4) + red); draw((-18.942629407869116,11.30491969191832)--(2.312195121951221,3.790243902439023), linewidth(0.4) + red); draw((-2,1)--(2.312195121951221,3.790243902439023), linewidth(0.4) + red); draw((2.312195121951221,3.790243902439023)--(6,12), linewidth(0.4) + red); draw((-18.942629407869116,11.30491969191832)--(-11,1), linewidth(0.4) + red); draw((-11,1)--(2.312195121951221,3.790243902439023), linewidth(0.4) + red); /* dots and labels */ dot((-11,1),dotstyle); label("$A$", (-10.843935095971819,1.379777826078773), NE * labelscalefactor); dot((-2,1),dotstyle); label("$B$", (-1.8398664483050566,1.379777826078773), NE * labelscalefactor); dot((6,12),dotstyle); label("$C$", (6.147613803657394,12.380716536736143), NE * labelscalefactor); dot((-3,12),linewidth(4pt) + dotstyle); label("$D$", (-2.8564548440093684,12.308103079900121), NE * labelscalefactor); dot((-6.639215718736752,3.8216839466997485),dotstyle); label("$K$", (-6.487127685810482,4.175395914265629), NE * labelscalefactor); dot((-3.5077820305878022,5.847905744913775),linewidth(4pt) + dotstyle); label("$L$", (-3.3647490418615242,6.13595924883823), NE * labelscalefactor); dot((-18.942629407869116,11.30491969191832),linewidth(4pt) + dotstyle); label("$E$", (-18.795108619516256,11.581968511539898), NE * labelscalefactor); dot((2.312195121951221,3.790243902439023),linewidth(4pt) + dotstyle); label("$F$", (2.4443275050202575,4.0664757290115965), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Too easy geo Rephrase the problem in terms of $F$. Restated problem wrote: Let $F$ be a point on circle $\omega$, such that $BF \parallel AC$, on line $AC$ choose $L$, let $E$ be the second intersection of $FL$ with $\omega$.Then let $K$ be the intersection of $AC$ with $BE$. Prove that $\angle KBA = \angle ADL$ Notice that $\angle DAL = \angle DAC = \angle ACB = \angle AEB$, thus we need to show that $\angle EAB = \angle ALD$, but this holds since : $$\angle EAB = 180 - \angle EFB = 180 - \angle LFB = 180 - \angle FLC = 180 - \angle CLD = \angle ALD$$thus we have that $EAB \sim ALD$, this implies that $\angle KBA = \angle ADL$, hence we are done