[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -19.30940365583981, xmax = 18.96633577975139, ymin = -13.870179855250106, ymax = 9.987792259311162; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); /* draw figures */ draw((-6.16,-1.66)--(7,4), linewidth(0.8) + blue); draw((-6.140828999727262,2.1572979519166013)--(-1.7733333333333337,0.22666666666666657), linewidth(0.8) + blue); draw((-6.140828999727262,2.1572979519166013)--(-1.754162333060595,4.043964618583268), linewidth(0.8) + blue); draw((-1.754162333060595,4.043964618583268)--(-1.7733333333333337,0.22666666666666657), linewidth(0.8) + blue); draw((-6.140828999727262,2.1572979519166013)--(-6.16,-1.66), linewidth(0.8) + blue); draw((-6.140828999727262,2.1572979519166013)--(-3.356645672941065,-0.45430201435003237), linewidth(0.8) + blue); draw((-1.754162333060595,4.043964618583268)--(2.572507074571529,-6.015923968773403), linewidth(0.8) + blue); draw((7,4)--(2.572507074571529,-6.015923968773403), linewidth(0.8) + blue); draw((-1.7733333333333337,0.22666666666666657)--(4.170555140458345,-2.400801777512361), linewidth(0.8) + blue); draw((-3.356645672941065,-0.45430201435003237)--(2.572507074571529,-6.015923968773403), linewidth(0.8) + blue); draw(circle((-3.5361502669936145,2.1442168937113957), 2.604711579971711), linewidth(1.2) + linetype("4 4") + qqwuqq); draw(circle((-1.7841609625778656,-1.9293130084284), 5.973352955177774), linewidth(1.2) + linetype("2 2") + red); draw((-1.754162333060595,4.043964618583268)--(4.170555140458345,-2.400801777512361), linewidth(0.8) + blue); draw((-1.754162333060595,4.043964618583268)--(-6.16,-1.66), linewidth(0.8) + blue); draw(circle((0.399586870619098,-2.8946286510533676), 3.803165349175038), linewidth(1.2) + linetype("2 2") + red); draw((-10.508324666121192,4.087929237166536)--(-6.140828999727262,2.1572979519166013), linewidth(0.8) + blue); draw((-10.508324666121192,4.087929237166536)--(7,4), linewidth(0.8) + blue); draw((-1.754162333060595,4.043964618583268)--(-3.356645672941065,-0.45430201435003237), linewidth(0.8) + blue); draw(circle((2.613333333333333,2.1133333333333337), 4.775181206567511), linewidth(1.2) + linetype("4 4") + qqwuqq); draw(circle((-6.140828999727262,2.1572979519166013), 4.775181206567511), linewidth(1.2) + linetype("4 4") + qqwuqq); /* dots and labels */ dot((-6.16,-1.66),dotstyle); label("$A$", (-6.007296901318203,-1.2117234275602247), NE * labelscalefactor); dot((7,4),dotstyle); label("$B$", (7.166079787837062,4.409489426769858), NE * labelscalefactor); dot((-1.7733333333333337,0.22666666666666657),linewidth(4pt) + dotstyle); label("$C$", (-1.5875646570739288,0.5904974875685042), NE * labelscalefactor); dot((-6.140828999727262,2.1572979519166013),dotstyle); label("$E$", (-5.964386879529425,2.6072685116411294), NE * labelscalefactor); dot((-1.754162333060595,4.043964618583268),linewidth(4pt) + dotstyle); label("$D$", (-1.5875646570739288,4.366579404981079), NE * labelscalefactor); dot((-3.356645672941065,-0.45430201435003237),linewidth(4pt) + dotstyle); label("$Z$", (-3.1752354632587654,-0.0960628610519639), NE * labelscalefactor); dot((2.572507074571529,-6.015923968773403),linewidth(4pt) + dotstyle); label("$K$", (2.7463475435927873,-5.674365693593267), NE * labelscalefactor); dot((4.170555140458345,-2.400801777512361),linewidth(4pt) + dotstyle); label("$T$", (4.334018349777624,-2.0699238633358097), NE * labelscalefactor); dot((0.2775344941629782,-0.6799099832298463),linewidth(4pt) + dotstyle); label("$F$", (0.42920636699870146,-0.3535229917846395), NE * labelscalefactor); dot((-0.3716561698038661,0.8295156594916505),linewidth(4pt) + dotstyle); label("$G$", (-0.21444395983298908,1.1912377926114137), NE * labelscalefactor); dot((-10.508324666121192,4.087929237166536),linewidth(4pt) + dotstyle); label("$H$", (-10.34120910198492,4.452399448558637), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Gotta hand it to the Greeks, they have phenomenal geo. Define $H=\overline{BD} \cap \overline{CE}$, $G=\overline{DK}\cap \overline{AB}$, $T=\overline{BK}\cap\overline{EC}$ and $F=\overline{DK}\cap\overline{EC}$. Claim 1: $ZEDC$ is cyclic. Proof: Since we have that $\omega = \angle AEZ = \angle ECA = \angle ECZ$, this implies that the triangles $CED$ and $ACE$ are isosceles. So this implies that $\angle ECD = 90- \frac{1}{2}\omega$, this implies that $\angle DEA = \angle ACD = \angle ZCD = 90 + \frac{1}{2}\omega$ and since we have that $\angle AEZ = \omega$, this implies that $\angle ZED = 90- \frac{1}{2}\omega$ Which implies that $\angle ZED + \angle ZCD = 180$, this implies that $ZCDE$ is cyclic. Claim 2: $EDTK$ is cyclic. Proof: Notice that we have that $\angle CGK = \angle CTK = 90 $, which implies that $CGTK$ is cyclic. Since we have that $AB \parallel ED$, we must have that: $$\frac{FC}{FG} = \frac{FE}{FD}$$by homothethy centered at $F$. By PoP we must have that: $$FC.FT=FG.FK \implies \frac{FC}{FG} = \frac{FK}{FT}$$this implies that: $$\frac{FE}{FD}=\frac{FK}{FT}$$this implies that $EF.FT = DF.FK$, thus by PoP we must have that $EDTK$ is cyclic. Claim 3: $BDCT$ is cyclic. Proof: Since we have that $AB=3.AC$, this implies that $BC=2.AC$, thus by a homothety centered at $H$ with coefficient $\frac{1}{2}$, we have that: $$EA=EC=EC$$this implies that $\angle HDC = 90$, which in turn implies that $\angle CDB = 90$, and since $\angle CTB = 90$, we have that $CDBT$ is cyclic. After all these claims we have that: $$\angle DEZ = \angle DCB = \angle DTB = \angle DEK$$this implies that $E$, $Z$ and $K$ are collinear points.

Another really nice Grecian Geometry Problem! We start off with the following characterization of $Z$. Claim : Quadrilateral $EZCD$ is cyclic. Proof : Simply note that, \[\measuredangle DEZ = \measuredangle DEA + \measuredangle AEZ = \measuredangle CAE + \measuredangle ECA = \measuredangle CEA = \measuredangle EAC = \measuredangle DCB \]which proves the claim. Now, let $P$ be the foot of the perpendicular from $B$ to $EC$ and $M$ the midpoint of $BC$. Further, let $K' = \overline{BP} \cap \overline{EZ}$. Now, note the followings. Claim : Quadrilateral $EDMC$ is a parallelogram. Proof : This is quite clear. We already have $ED \parallel CM$ by construction. Further note that, \[CM = \frac{CB}{2}=\frac{2AC}{2}=AC=ED\]which implies the desired claim. Claim : Quadrilateral $DCPB$ is cyclic. Proof : We note that since $BM=CM=AC=EC=DM$, we have that $M$ must be the circumcenter of $\triangle DCB$. This implies that $\triangle DCB$ must be right and thus we have that \[\measuredangle CPB = 90^\circ = \measuredangle CDB \]and thus $DCPB$ must be cyclic. Now we are at the final stretch. Note that, \[\measuredangle DEK' = \measuredangle DEZ = \measuredangle DCB = \measuredangle DPB\]from which it is clear that $DEK'P$ is also cyclic. But this means, $\measuredangle K'DE = \measuredangle K'PE = 90^\circ$. Thus, $K'D \perp DE$ which implies that $K'D \perp AB$ as well. Thus, we must have $K'=K$ and it follows that the line passing through point $B$ and perpendicular on side $EC$, and the line passing through point $D$ and perpendicular on side $AB$, intersect on point lying on line $EZ$.