[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10.455170115890374cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -2.1486136382007968, xmax = 3.07897141974439, ymin = -1.3679916560420065, ymax = 1.3627420138882473; /* image dimensions */ pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); pen zzttqq = rgb(0.6,0.2,0.); pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen ffqqff = rgb(1.,0.,1.); pair A = (-0.5509314319221846,0.8345505121442148), B = (-0.8945276279430447,-0.44701266519707206), C = (0.8280646749728561,-0.5606325838390935), D = (-0.34996828208177905,-0.4829310821479518), F = (-0.7532013792947415,0.08011341581590979), K = (-0.8363980305018978,-0.23019783296964078), L = (-0.6700047280875853,0.39042466460146047); draw(A--B--C--cycle, linewidth(1.2) + zzttqq); draw(arc((-0.0018553322945480955,0.2790292247536388),0.06321142754468183,170.53457342483057,194.82862432943026)--(-0.0018553322945480955,0.2790292247536388)--cycle, linewidth(2.) + qqwuqq); draw(arc((-0.0018553322945480955,0.2790292247536388),0.06321142754468183,-165.17137567056977,-140.8773247659701)--(-0.0018553322945480955,0.2790292247536388)--cycle, linewidth(2.) + qqwuqq); draw(arc(D,0.06321142754468183,98.67280774998346,125.60893532905226)--(-0.34996828208177905,-0.4829310821479518)--cycle, linewidth(2.) + ffqqff); draw(arc(D,0.06321142754468183,125.60893532905226,152.54506290812103)--(-0.34996828208177905,-0.4829310821479518)--cycle, linewidth(2.) + ffqqff); /* draw figures */ draw(circle((0.,0.), 1.), linewidth(1.2)); draw(A--B, linewidth(1.2) + zzttqq); draw(B--C, linewidth(1.2) + zzttqq); draw(C--A, linewidth(1.2) + zzttqq); draw(circle((-0.3204828005355085,-0.03590151051622459), 0.4480009280517853), linewidth(1.2)); draw(L--(-0.0018553322945480955,0.2790292247536388), linewidth(1.2)); draw((-0.0018553322945480955,0.2790292247536388)--F, linewidth(1.2)); draw(K--D, linewidth(1.2)); draw(D--A, linewidth(1.2)); draw((-0.0018553322945480955,0.2790292247536388)--B, linewidth(1.2)); draw(D--F, linewidth(1.2)); /* dots and labels */ dot(A,xdxdff); label("$A$", (-0.5367222358114101,0.8665323076624951), NE * labelscalefactor,xdxdff); dot(B,xdxdff); label("$B$", (-0.8812245159299261,-0.41665967149454536), NE * labelscalefactor,xdxdff); dot(C,xdxdff); label("$C$", (0.8412868846626537,-0.5304402410749726), NE * labelscalefactor,xdxdff); dot(D,linewidth(4.pt) + uuuuuu); label("$D$", (-0.3660513814407692,-0.5588853834700794), NE * labelscalefactor,uuuuuu); dot((-0.0018553322945480955,0.2790292247536388),linewidth(4.pt) + uuuuuu); label("$E$", (0.010056612450087633,0.30395060251482714), NE * labelscalefactor,uuuuuu); dot(F,linewidth(4.pt) + uuuuuu); label("$F$", (-0.7927285173673715,0.10799517712631355), NE * labelscalefactor,uuuuuu); dot(K,linewidth(4.pt) + uuuuuu); label("$K$", (-0.8907062300616284,-0.17961681820198863), NE * labelscalefactor,uuuuuu); dot(L,linewidth(4.pt) + uuuuuu); label("$L$", (-0.7263565184454556,0.40508888658631803), NE * labelscalefactor,uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] the first part is a simple angle chase and it is trivial.for the second part we have \[\frac{LF}{FB}=\frac{LE}{EB}=\frac{AE}{AB} \quad \frac{KF}{FA}=\frac{KD}{AD}=\frac{BD}{AB}\]from these equations we need to prove that $AE.FB=BD.FA$ which is trivial since $AE=AF,BF=BD$ really easy P6

@above: Each day of the Thailand MO has 5 problems; thus, P6 is the first problem on the second day.

a. It suffices to show $\angle AEL=\angle ABE$. We have $\angle AEL=\angle AEF-\angle LEF$. But since $\angle AFE=\angle FBE+\angle BEF$, $\angle ABE=\angle FBE=\angle ABE$ (using the facts that $\angle AFE=\angle AEF$ and $\angle LEF=\angle BEF$). b. From the similarity, we have that $AL=\frac{AE^2}{AB}=\frac{AF^2}{AB}$. We can now compute \[FL=AF-AL=AF\left(1-\frac{AF}{AB}\right)=\frac{AF\cdot BF}{AB}\]By symmetry, we are done.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9cm); 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 = -49.21949553199036, xmax = 64.9636554566552, ymin = -34.02484486610404, ymax = 33.067085028531864; /* image dimensions */ /* draw figures */ draw((2.6000327484570454,17.359875625216375)--(-7.511428042782372,-7.834043441223872), linewidth(0.8) + blue); draw((-7.511428042782372,-7.834043441223872)--(24.122192309698484,-12.127177631917725), linewidth(0.8) + blue); draw((24.122192309698484,-12.127177631917725)--(2.6000327484570454,17.359875625216375), linewidth(0.8) + blue); draw(circle((4.8509844546460865,-0.6384367796172477), 8.79275797832267), linewidth(0.8) + red); draw((-3.3090954828131887,2.636572881380963)--(3.6685215651774565,-9.351322316589858), linewidth(0.8) + blue); draw((2.6000327484570454,17.359875625216375)--(3.6685215651774565,-9.351322316589858), linewidth(0.8) + blue); draw((3.6685215651774565,-9.351322316589858)--(-5.764934725932467,-3.4824455679553727), linewidth(0.8) + blue); draw((11.953169304292906,4.5453420762758325)--(-3.3090954828131887,2.636572881380963), linewidth(0.8) + blue); draw((11.953169304292906,4.5453420762758325)--(-7.511428042782372,-7.834043441223872), linewidth(0.8) + blue); draw((11.953169304292906,4.5453420762758325)--(-0.8532562396939101,8.755591330717298), linewidth(0.8) + blue); draw(circle((2.617776652458045,-2.2684217146066747), 11.55754874365122), linewidth(0.8) + red); draw(circle((5.493876656561384,4.0986642472212385), 13.573284782112998), linewidth(0.8) + red); /* dots and labels */ dot((2.6000327484570454,17.359875625216375),dotstyle); label("$A$", (2.8719027621760453,18.141182938512753), NE * labelscalefactor); dot((-7.511428042782372,-7.834043441223872),dotstyle); label("$B$", (-7.203081148586798,-7.083591593619546), NE * labelscalefactor); dot((24.122192309698484,-12.127177631917725),dotstyle); label("$C$", (24.43983128225354,-11.412103199725088), NE * labelscalefactor); dot((3.6685215651774565,-9.351322316589858),linewidth(4pt) + dotstyle); label("$D$", (3.9913454189274726,-8.725440823521648), NE * labelscalefactor); dot((11.953169304292906,4.5453420762758325),linewidth(4pt) + dotstyle); label("$E$", (12.275221078888032,5.155648120196125), NE * labelscalefactor); dot((-3.3090954828131887,2.636572881380963),linewidth(4pt) + dotstyle); label("$F$", (-3.0238285633814703,3.2152808484936406), NE * labelscalefactor); dot((-5.764934725932467,-3.4824455679553727),linewidth(4pt) + dotstyle); label("$K$", (-5.48660240823461,-2.904339008414195), NE * labelscalefactor); dot((-0.8532562396939101,8.755591330717298),linewidth(4pt) + dotstyle); label("$L$", (-0.5610547185283309,9.334900705401475), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Nice exercise $\color{black}\rule{25cm}{1pt}$ Solution for part a. To show that the statement holds true we want to show that $AC$ is tangent to the circumcircle of $BEL$. But this automatically pops up since we have that $\angle FBE+\angle EBF = \angle AFE = \angle AEF = \angle FEL + \angle LEA= \angle FEB + \angle LEA$. From this we have that $\angle EBL = \angle LEA$, thus we have the tangency. Automatically we see that $BC$ is tangent to the circumcircle of $ADK$ by doing a similar procedure to the one above. Solution for part b. From PoP we have that $AL.AB=AE^2$ and $BK.BA=BD^2$, this gives us $AL=\frac{AE^2}{AB}$ and $BK=\frac{BD^2}{BA}$. Let's denote with $a=BC$, $b=CA$, $c=AB$ and with $s$ we shall denote the half-perimeter. We have that $FL=AF-AL$ and $KF=BF-BK$, thus we need to show that $AF-AL=BF-BK$. Since we have that $AF=s-a$ and that $BD=s-b$ the relation gets turned into: $$s-a-\frac{(s-a)^2}{c}=s-b-\frac{(s-b)^2}{c}$$which turns into the following: $$4ac+(b+c-a)^2=4bc+(c+a-b)^2$$which when expanded gives us that $LHS=RHS$, thus we must have that $FL=FK$

$L_1$ be a point on $AB$ such that $\triangle ALE\sim\triangle AEB$.Then $AE^2=AL_1\times AB$.So $A$ is the centre of $E$-appolonian circle $\omega$ of $L_1EB$.Also $AE=AF$ implies $F\in \omega$.So $EF$ bisects $L_1EB$ i.e. $L_1\equiv L$ $\blacksquare$ We have $$\frac{LF}{FB}=\frac{LE}{BE}=\frac{AE}{AB}\implies LF=\frac{AE\times FB}{AB}$$So,$LF=\frac{(s-a)(s-b)}{c}$. Similarly $FD$ bisects $ADK$ and $BDK\sim BAD$ implies $FK=\frac{(s-a)(s-b)}{c}$.We are done.$\blacksquare$