Clue please

parmenides51 wrote: Given quadrilateral $ABCD$. Suppose $E, F, G, H$ are respectively the midpoint of the sides $AB, BC, CD, DA$. The line passing through $G$ and perpendicular on $AB$ intersects the line passing through $H$ and perpendicular on $BC$ at point $K$. Prove that $\angle EKF = \angle ABC$. I think the condition " $ABCD$ is cyclic" is missing. @below thanks for checking.

yes , you are correct, the word cyclic is missing, here is my source using google translate when I posted that thanks a lot for noticing

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(8.67546200478607cm); 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 = -2.500636965306116, xmax = 6.174825039479954, ymin = 0.7919197029191294, ymax = 5.486249450294183; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw((-0.7086194720093624,3.4583613180592248)--(3.077206090531799,4.565479556534677)--(3.1011825897675136,1.7040543618155364)--(-0.335203848023057,1.862954905006177)--cycle, linewidth(0.8) + zzttqq); draw((0.7986933547457992,3.7815231982467803)--(0.9070591904365475,3.8132134573828433)--(0.8753689313004842,3.9215792930735915)--(0.767003095609736,3.8898890339375285)--cycle, qqwuqq); draw((3.0919671811134455,2.803848049090182)--(2.9790666411721825,2.8029020311007518)--(2.9800126591616127,2.6900014911594887)--(3.0929131991028758,2.690947509148919)--cycle, qqwuqq); draw((1.087109041941943,1.910211459613351)--(0.9743250511734431,1.9154266581135073)--(0.9691098526732869,1.8026426673450076)--(1.0818938434417866,1.7974274688448513)--cycle, qqwuqq); draw((-0.41383487892575366,2.1989034912702996)--(-0.30390144478164216,2.224634152481147)--(-0.3296321059924893,2.3345675866252584)--(-0.43956554013660076,2.308836925414411)--cycle, qqwuqq); /* draw figures */ draw(circle((1.444835417606714,3.1209885208911516), 2.179722084551239), linewidth(0.8)); draw((-0.7086194720093624,3.4583613180592248)--(3.077206090531799,4.565479556534677), linewidth(0.8) + zzttqq); draw((3.077206090531799,4.565479556534677)--(3.1011825897675136,1.7040543618155364), linewidth(0.8) + zzttqq); draw((3.1011825897675136,1.7040543618155364)--(-0.335203848023057,1.862954905006177), linewidth(0.8) + zzttqq); draw((-0.335203848023057,1.862954905006177)--(-0.7086194720093624,3.4583613180592248), linewidth(0.8) + zzttqq); draw((xmin, -3.419531384248895*xmin + 6.512680191191076)--(xmax, -3.419531384248895*xmax + 6.512680191191076), linewidth(0.8)); /* line */ draw((xmin, 0.008379215811744521*xmin + 2.665031321966643)--(xmax, 0.008379215811744521*xmax + 2.665031321966643), linewidth(0.8)); /* line */ draw((1.1842933092612182,4.011920437296951)--(1.122447262526732,2.674436549816656), linewidth(0.8)); draw((3.089194340149656,3.1347669591751064)--(1.122447262526732,2.674436549816656), linewidth(0.8)); draw((xmin, 21.62602070949355*xmin-21.5996311949008)--(xmax, 21.62602070949355*xmax-21.5996311949008), linewidth(0.8)); /* line */ draw((xmin, 0.23405673998245938*xmin + 2.411720202747413)--(xmax, 0.23405673998245938*xmax + 2.411720202747413), linewidth(0.8)); /* line */ /* dots and labels */ dot((1.444835417606714,3.1209885208911516),linewidth(2.pt) + dotstyle); label("$O$", (1.464528184120646,3.1444069459347457), NE * labelscalefactor); dot((3.077206090531799,4.565479556534677),linewidth(3.pt) + dotstyle); label("$B$", (3.1251074144846056,4.624025619143663), NE * labelscalefactor); dot((-0.7086194720093624,3.4583613180592248),linewidth(3.pt) + dotstyle); label("$A$", (-0.5952287458500344,3.5648741228538263), NE * labelscalefactor); dot((3.1011825897675136,1.7040543618155364),linewidth(3.pt) + dotstyle); label("$C$", (3.157041630453143,1.6062422101168428), NE * labelscalefactor); dot((-0.335203848023057,1.862954905006177),linewidth(3.pt) + dotstyle); label("$D$", (-0.3131431714612849,1.8936501538336827), NE * labelscalefactor); dot((1.1842933092612182,4.011920437296951),linewidth(3.pt) + dotstyle); label("$E$", (1.256955780325151,4.123722902303237), NE * labelscalefactor); dot((3.089194340149656,3.1347669591751064),linewidth(3.pt) + dotstyle); label("$F$", (3.1091403065003367,3.165696423247104), NE * labelscalefactor); dot((1.3829893708722283,1.7835046334108569),linewidth(3.pt) + dotstyle); label("$G$", (1.4059821215116601,1.8138146139123383), NE * labelscalefactor); dot((-0.5219116600162097,2.660658111532701),linewidth(3.pt) + dotstyle); label("$H$", (-0.4994260979444214,2.692005553047127), NE * labelscalefactor); dot((1.122447262526732,2.674436549816656),linewidth(4.pt) + dotstyle); label("$K$", (1.1451860244352692,2.7186173996875755), NE * labelscalefactor); dot((0.767003095609736,3.8898890339375285),linewidth(3.pt) + dotstyle); label("$G'$", (0.7726201714689962,4.001308407757176), NE * labelscalefactor); dot((3.0929131991028758,2.690947509148919),linewidth(3.pt) + dotstyle); label("$H'$", (3.114462675828426,2.723939769015665), NE * labelscalefactor); dot((1.0818938434417866,1.7974274688448513),linewidth(3.pt) + dotstyle); label("$E'$", (0.9163241433274157,1.8989725231617725), NE * labelscalefactor); dot((-0.43956554013660076,2.308836925414411),linewidth(3.pt) + dotstyle); label("$F'$", (-0.41959055802307715,2.3407291773932117), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $E', F', G', H'$ be the points of intersection from the perpendiculars through $E, F, G, K$ and $DC, AD, AB, BC$, then we have quadrilaterals $F'DE'K$ and $H'BG'K$ are cyclic, $$\measuredangle EKF=\measuredangle E'KF'=\measuredangle E'DF'=\measuredangle CDA=\measuredangle ABC.$$