Well the only problem here is to get rid of things with the point $K$ so that we can angle chase pretty much anything.

EDIT: Oops! I read the problem statement wrong and confused $C$ with $D$. Well at least we know that the statement holds even in that case

A very cute problem. Let $H$ be the intersection of $AD$ and $CB$.We will proceed proving that $K,H,M$ are collinear. In cyclic quadrilateral $AOKC$ we have $\angle AKO=\angle ACO$.Simillary in the quadrilateral $BOKD$ we get $\angle BKO=\angle BDO$. In the cercle with diameter $AB$. We get $\angle ACO=90-\frac{\angle COA}{2}$ and $\angle BDO=90-\frac{BOD}{2}$. So we get $\angle AKB=\angle AKO+\angle OKB=180-(\frac{\angle COA+\angle BOD}{2})=180-\angle CHA=\angle AHB$ Therefore $\angle AKB=\angle AHB$ so the quadrilateral $AKHB$ is cyclic. As a consequence we get $\angle CKH=360-\angle AKH-\angle CKA=(180-\angle AKH)+(180-\angle CKA)=\angle HBA-(180-\angle COA)$. So we get $\angle CKH+\angle CDH=180$ hence the quadrilateral $AKHD$ is cyclic. The 3 radical axis of circumcercles $AKHB,CKHD,ABCD$ intersect in the radical center. But $CD,AB$ intersect at $M$ so we get that $K,H,M$ are collinear. Now let us prove that $HM\perp SO$ where $S$ is the intersection of $AC,DB$. We conclude that $CB,AD$ are altitudes in $\triangle ABS$ because $AB$ is diameter,and $SO$ is the median. Let $SU$ be the third altitude and let $K'$ be the projection of $H$ on $SO$. It is easy to observe that the penthagon $HK'CDS$ is cyclic because $\angle SCB=\angle SK'H=\angle SDH=90$ Also it is easy to see that the quadrilateral $HK'OU$ is cyclic $\angle HK'O=\angle HUO$. But the cercle of Euler of the $\triangle SBA$ contains the points $D,C,U,O$. So by the monge theorem or the radical axis theorem we get $H,K',M$ are collinear so $K=K'$. So we are done.

http://web.mit.edu/yufeiz/www/cyclic_quad.pdf here u can find some interesting stuff about... I saw it on forum, actually it's Russia 1995 and it was also used in Iranian MO and Romanian TST, but I couldn't find any link now...

See problem 58 of the geometry marathon: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=331763&p=1896287#p1896287

See this as well: $AC \cap BD={E}$, $AD \cap BC = {H}$; $D$ and $E$ having equal powers w.r.t. $\odot (OAC)$, $\odot (OBD)$, $K$ is on $OE$. Next, it is well known that for 3 points (let them be $C, O, D$) on the sides $AE, AB, BE$ of any $\triangle ABE$, the circles $\odot (AOC)$, $\odot (BOD)$, $\odot (CDE)$ have a common point, hence $ECKD$ is cyclic, and easy to see, $EH$ is its circumcircle diameter. $ABDC$ being cyclic, $\triangle ABE \sim \triangle DCE$ and, $EO$ being median in $\triangle EAB$, it is symmedian in $\triangle EDC$, hence $OC$, $OD$ are tangent to $\odot (ECKHD)$, or $\angle OCK= \angle CDK$ and $\angle ODK= \angle DCK$, but $\angle OCK=\angle OAK$ and $\angle ODK=\angle OBK$, or $\angle OAK+\angle OBK=\angle CDK+\angle DCK=\angle CED$, hence $\angle AKB=\angle AHB=180^\circ -\angle CED$, that is, $ABHK$ is cyclic, but $M$ has equal powers w.r.t. $\odot (ABHK)$ and $\odot (ECKHD)$, or $M$ belongs to their radical axis, i.e. $M, H, K$ are collinear. As $EH$ is a diameter of $\odot (ECKHD)$, it follows that $HK \perp EK$, or $MK \perp OK$. Best regards, sunken rock

Invert the whole figure w.r.t $ \odot (ABC) $, we note that $\odot (AOD)$ maps to $AC$ and $\odot (BOC)$ maps to $ BD $. Denoting the image of a point under that inversion $X'$ , we have $K'$ is the intersection of $AC$ and $BD$. $AB$ is mapped to itself and $CD$ is mapped to $\odot( COD)$. As $M$ is the intersection of $AB$ and $CD$. $M'$ is the intersection of $\odot COD$ and $AB$. But we note that $ \odot( COD)$ is the nine point circle of $\triangle AKB$, and as it intersects $AB$ at $O$ and $M'$ , where $O$ is the mid-point of $\overline{AB}$ , it follows that $M'$ is the foot of altitude dropped from $K'$, that's $\angle K'M'O = \frac{\pi}{2}$, but By a property of inversion $\angle MKO = \angle K'MO = \frac{\pi}{2}$. So we are done.

Nice problem! Another Inversion approach using Brocard to finish off. Invert the whole figure about the given semicircle. The essential mappings have been mentioned by Nanas already, so I won't also include them here. We'll show that $AD$, $BC$ and $K'M'$ are concurrent by Ceva. Let the radius of the circumcircle be $r$, then \[ |AM'|=\frac{r^2}{|OA| \cdot |OM|} \cdot |AM| = \frac{r}{|OM|} \cdot |AM| \quad \text{and} \quad |M'B|=\frac{r}{|OM|} \cdot |BM|. \]We also have \[ |DK'| = \frac{r^2}{|OD| \cdot |OK|} \cdot |DK| = \frac{r}{|OK|} \cdot |DK| \quad \text{and} \quad |K'C| = \frac{r}{|OK|} \cdot |DK|. \]Both are just an application of the Inversion Distance Formula (which follows from simple similarity). Plugging those in yields \[ \frac{|AM'|}{|M'B|} \cdot \frac{|BD|}{|DK'|} \cdot \frac{|K'C|}{|CA|} = \frac{|AM|}{|BM|} \cdot \frac{|BD|}{|CA|}. \]But that's equal to $1$ as $\triangle BMD \sim \triangle AMC$ by $\angle DMB = \angle CMA$ and $\angle MBD = \angle ACM$ as $ABCD$ is cyclic. The similarity gives $\tfrac{|AM|}{|CA|} = \tfrac{|BM|}{|BD|}$ which is just what we wanted. Therefore $AD$, $BC$ and $K'M'$ intersect in a point which we'll call $S$. By Brocard's Theorem $O$ is the orthocenter of $\triangle SMK'$. Thus $MO \perp K'M'$. Note that we needed that $K', S, M'$ to be collinear for that. Thus \[ 90^{\circ} = \angle OM'K' = \angle OKM, \]phew! $\blacksquare$

Super-quick solution by inversion! Inequalities Master wrote: Consider a semicircle of center $O$ and diameter $AB$. A line intersects $AB$ at $M$ and the semicircle at $C$ and $D$ s.t. $MC>MD$ and $MB<MA$. The circumcircles od the $AOC$ and $BOD$ intersect again at $K$. Prove that $MK\perp KO$. Invert around the semicircle! Let $X^*$ denote image of $X$ under inversion.

[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(0.6) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = blue; /* point style */ real xmin = -7.77, xmax = 13.41, ymin = -2.05, ymax = 10.61; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(shift((-0.07,-0.01))*xscale(5.140972670613997)*yscale(5.140972670613997)*arc((0,0),1,1.114563272484713,181.11456327248473), linewidth(2) + wrwrwr); draw((2.0544204131285877,9.55076727464112)--(5.07,0.09), linewidth(1) + wrwrwr); draw((-5.21,-0.11)--(11.36926798405717,0.21255385182990602), linewidth(1) + wrwrwr); draw((-5.21,-0.11)--(2.0544204131285877,9.55076727464112), linewidth(1) + wrwrwr); draw((-1.4008170322545879,4.955735194979804)--(11.36926798405717,0.21255385182990602), linewidth(1) + wrwrwr); draw(circle((-2.6791429333400174,1.951946773676892), 3.264484950086202), linewidth(1) + linetype("2 2") + wrwrwr); draw(circle((1.0277160171656186,2.945384972412403), 3.152662492164837), linewidth(1) + wrwrwr); draw(circle((2.485127701233374,0.8044361566045767), 2.6817874305759175), linewidth(1) + linetype("2 2") + wrwrwr); draw((2.0544204131285877,9.55076727464112)--(-0.07,-0.01), linewidth(2) + wrwrwr); draw((0.515349697261791,2.6243148442826065)--(2.2395533767388622,0.03493294507274051), linewidth(1) + wrwrwr); draw((2.2395533767388622,0.03493294507274051)--(2.0544204131285877,9.55076727464112), linewidth(1) + wrwrwr); draw((0.515349697261791,2.6243148442826065)--(11.36926798405717,0.21255385182990602), linewidth(1) + wrwrwr); /* dots and labels */ dot((-5.21,-0.11),dotstyle); label("$A$", (-5.75,0.09), NE * labelscalefactor); dot((5.07,0.09),dotstyle); label("$B$", (5.15,0.29), NE * labelscalefactor); dot((-1.4008170322545879,4.955735194979804),dotstyle); label("$C$", (-1.75,5.15), NE * labelscalefactor); dot((11.36926798405717,0.21255385182990602),linewidth(4pt) + dotstyle); label("$M$", (11.45,0.37), NE * labelscalefactor); dot((4.179760149136906,2.8829463656984515),linewidth(4pt) + dotstyle); label("$E$", (4.25,3.05), NE * labelscalefactor); dot((2.0544204131285877,9.55076727464112),linewidth(4pt) + dotstyle); label("$K'$", (2.13,9.71), NE * labelscalefactor); dot((-0.07,-0.01),linewidth(4pt) + dotstyle); label("$O$", (0.01,0.15), NE * labelscalefactor); dot((0.515349697261791,2.6243148442826065),linewidth(4pt) + dotstyle); label("$K$", (0.59,2.79), NE * labelscalefactor); dot((2.2395533767388622,0.03493294507274051),linewidth(4pt) + dotstyle); label("$M'$", (2.31,0.19), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We invert about $O$ with radius $OA$. We have the following as a result of this inversion : $\odot(OCA) \mapsto AC$ $\odot(ODB) \mapsto BD$ $K' = AC \cap BD$ $CD \cap AB = M \Longrightarrow C'D' \cap A'B' = M' \Longrightarrow M' = \odot(OCD) \cap AB$ Because, $\odot(OCD)$ is the Nine Point Circle of $\triangle K'AB$, $M'$ is the foot of the perpendicular from $K'$ to $AB$. Thus, $\angle MM'K' = 90 \Longrightarrow \angle MKO = 90$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -0.706075850192391, xmax = 5.072838009704374, ymin = -0.7675713980837071, ymax = 2.6049666123404194; /* image dimensions */ /* draw figures */ draw(shift((1.734541780261432,0.2464552288324917))*xscale(1.1888167648276609)*yscale(1.1888167648276609)*arc((0,0),1,-1.5287238730419253,178.47127612695806), linewidth(1)); draw((2.9229354168685604,0.21473985705490742)--(0.5461481436543035,0.278170600610076), linewidth(0.8)); draw((1.7662154856733812,1.4348499766838878)--(4.331115913656617,0.17715890188159983), linewidth(0.8)); draw((4.331115913656617,0.17715890188159983)--(2.9229354168685604,0.21473985705490742), linewidth(0.8)); draw(circle((1.1562020918690559,0.8564889002169811), 0.8406056598623577), linewidth(0.4)); draw(circle((2.3346381735238486,0.451658086990116), 0.6342112378400585), linewidth(0.4)); draw((4.331115913656617,0.17715890188159983)--(1.987336064771597,0.9823228959409812), linewidth(0.8)); draw((0.5461481436543035,0.278170600610076)--(2.324675250619426,1.9642952541971175), linewidth(0.8)); draw((2.324675250619426,1.9642952541971175)--(2.9229354168685604,0.21473985705490742), linewidth(0.8)); draw((1.734541780261432,0.2464552288324917)--(2.324675250619426,1.9642952541971175), linewidth(0.8)); draw(circle((2.0223502529347988,0.8334038758626591), 0.6537143345476831), linewidth(0.8)); draw((2.324675250619426,1.9642952541971175)--(2.278442844885253,0.23193981580485437), linewidth(0.8)); /* dots and labels */ dot((0.5461481436543035,0.278170600610076),dotstyle); label("$A$", (0.44067736888087333,0.0753419625695368), NE * labelscalefactor); dot((2.9229354168685604,0.21473985705490742),dotstyle); label("$B$", (2.8876862064309097,0.01635433961907339), NE * labelscalefactor); dot((1.734541780261432,0.2464552288324917),dotstyle); label("$O$", (1.625785222827202,0.07086911607211774), NE * labelscalefactor); dot((1.7662154856733812,1.4348499766838878),dotstyle); label("$C$", (1.7273886579985123,1.4988463813445545), NE * labelscalefactor); dot((2.6546937164212374,0.9991873954871838),dotstyle); label("$D$", (2.670976936684781,1.0428539595870756), NE * labelscalefactor); dot((4.331115913656617,0.17715890188159983),dotstyle); label("$M$", (4.327899894952057,0.18019419803909336), NE * labelscalefactor); dot((1.987336064771597,0.9823228959409812),dotstyle); label("$K$", (1.7854058338550643,1.0022209715096764), NE * labelscalefactor); dot((2.324675250619426,1.9642952541971175),dotstyle); label("$K^*$", (2.3413982556125434,2.0090161205385657), NE * labelscalefactor); dot((2.278442844885253,0.23193981580485437),dotstyle); label("$M^*$", (2.2962504910821,0.06989866897820644), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]Let $X^*$ denote the inversion image of $X$. $$\mathbf{SOLUTION:}$$Inverting about $O$ with radius $AO$,we see:$\newline$ $(AOC)\mapsto AC \newline (BOD)\mapsto BD \newline CD\mapsto (COD)\newline$ Since $K=(COA)\cap (DOB)-\{O\}$ and $M=CD\cap AB$,we get: $\newline K^*=AC\cap BD\newline M^*=(COD)\cap AB-\{O\}$$\newline$ Now, $\angle MKO=90^{\circ} \Longleftrightarrow \angle K^*M^*O=90^{\circ}$ $\newline$ Since $\angle ADB=90^{\circ}$,$\angle BCA=90^{\circ}$ and $O$ is the midpoint of $AB$ it follows that $(CDO)$ is the nine-point circle of $\triangle K^*AB$, hence $\angle K^*M^*O=90^{\circ}\Longrightarrow \angle MKO=90^{\circ}$.$\blacksquare$

Why is $K^*$ the intersection of $BD$ and $AC?$

consider an inversion with respect to semicircle (AB) with center O Now K^,C^,A^ and A^,O,M^,B^ and K^,D^,B^ are collinear thus K^,A^,B^ is a triangle and C^,O,M^,D^ is cyclic also <B^C^A^= 90 =<A^D^B^ and O is the midpoint of AB therefore C^,O,M^,D^ is the nine point cirlce of triangle A^,B^,K^ so <K^M^A^ =90 =<K^M^0 =<OM^K^=<OKM=<MKO threfore <MKO=90 (as desired)

It's been a while since I tried to solve something... Let $X = AC \cap BD$, $H$ - orthocenter of $\triangle ABX$ (also $AD \cap BC$). We have $X \in OK$ (Radical Center of $(O),(OAC),(OBD)$). Also $XCDK$ - cyclic (by Miquel on $\triangle XAB$). Since $H$ is also on this circle, $HK \perp KO$. Also by Brocard $H$ - orthocenter of $\triangle OXM$, so $M \in HK$. Hence proved.

Consider $S=AC \cap BD, T=AD \cap BC$. 1) $O,K,S$ collinear. By Power of Point on the semi-cricle, $SC \cdot SA = SD \cdot SB \Rightarrow S \in $ Radical Axis of $(AOC), (BOD.)$ The result follows. 2) By Brocard' Theorem on $ABCD,$ we have that $MT \perp OS, \Rightarrow MT \perp OK.$ Notice that $T$ is the orthocenter of $\triangle{ABS}.$ 3) $K \in (SCTD).$ Notice that $ \angle{KCS}=180-\angle{KCA}=\angle{KOA}=180-\angle{KOB}=\angle{KDB}=180-\angle{KDS}.$ 4) Since $K$ also lies on the $S$-median of $\triangle{ABS}$, we have that $K$ is the $S$-Humpty Point of this triangle. It is a well-known fact about the A- Humpty that $CD, AB,$ and $KT$ must concur, that $M, K, T $ are collinear. Thus, by (2), the result follows.

Here is a proof that I finish off using Menelaus, but I feel like there has to be a simpler way to finish using projective geometry (perhaps Brianchon's?). Let $P$ and $Q$ be the centers of the circumcircles $O_1$ and $O_2$ of $\triangle AOC$ and $\triangle BOD$. Let $OP$ meet $O_1$ at $R$, and let $OQ$ meet $O_2$ at $S$. We have $PQ \perp OK$ as the line connecting two circle centers is perpendicular to their radical axis. Since $OP=PR$ and $OQ=QS$, $RS \parallel PQ$, so $RS \perp OK$. Since $RO$ is a diameter of $O_1$, $\angle RAO= \angle RCO = 90$, so $RA$ and $RC$ are tangent to circle $O$. Similarly, $\angle SBO=90$, and $SB$ and $SD$ are tangent to circle $O$. So $AR=RC$ and $BS=SD$. Let $RC$ meet $SD$ at $E$, and let $RS$ meet $AB$ at $M'$. Then $\frac{M'S}{M'R}=\frac{BS}{AR}$. We have $\frac{M'S}{M'R} \cdot \frac{RC}{CE} \cdot \frac{ED}{DS}= \frac{BS}{AR} \cdot \frac{AR}{CE} \cdot \frac{CE}{BS} =1$. Thus, by the converse of Menelaus, $C$,$D$, and $M'$ are collinear, so $M'=M$ and $OK\perp MK$.

Inequalities Master wrote: Consider a semicircle of center $O$ and diameter $AB$. A line intersects $AB$ at $M$ and the semicircle at $C$ and $D$ s.t. $MC>MD$ and $MB<MA$. The circumcircles od the $AOC$ and $BOD$ intersect again at $K$. Prove that $MK\perp KO$.

Invert around the given semicircle. Observe that $(OAC)$ becomes $AC$ and $(OEB)$ becomes $EB$. Since the two circles intersect at $k$, $k'$ must be the intersection of $AC$ and $BE$. Further, since $M, E, C$ are collinear, $M'$ is the intersection of $(CEO)$ and $OM$. Since we want to show that $OM$ is the diameter of $(KOM)$ it suffices to show that $K'M' \perp OM$. But, observe $(OECM)$. Since $CO=AO=OB=OE$, we have that $BC$ and $AE$ are altitudes of $\triangle K'AB$. Hence, $(COM'E)$ is the nine point circle, and $M'$ must be the altitude from $k'$.

Inequalities Master wrote: Consider a semicircle of center $O$ and diameter $AB$. A line intersects $AB$ at $M$ and the semicircle at $C$ and $D$ s.t. $MC>MD$ and $MB<MA$. The circumcircles od the $AOC$ and $BOD$ intersect again at $K$. Prove that $MK\perp KO$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -25.174219652847135, xmax = 30.87847504322665, ymin = -10.49872590800251, ymax = 24.118684722292507; /* image dimensions */ pen qqttzz = rgb(0,0.2,0.6); pen qqzzcc = rgb(0,0.6,0.8); pen ttqqtt = rgb(0.2,0,0.2); pen yqqqyq = rgb(0.5019607843137255,0,0.5019607843137255); pen ffwwqq = rgb(1,0.4,0); /* draw figures */ draw(circle((0,0), 5), linewidth(0.8) + qqttzz); draw((-4,3)--(13.30857214203163,0), linewidth(0.4) + qqzzcc); draw((13.30857214203163,0)--(-5,0), linewidth(0.4) + qqzzcc); draw(circle((-2.5,0.8333333333333334), 2.6352313834736494), linewidth(0.4) + linetype("4 4") + ttqqtt); draw(circle((2.5,0.3781731346741374), 2.5284412035460244), linewidth(0.4) + linetype("4 4") + yqqqyq); draw((1.878488521022031,0)--(1.8784885210220088,20.635465563066035), linewidth(0.4) + linetype("4 4") + ffwwqq); draw((1.8784885210220088,20.635465563066035)--(13.30857214203163,0), linewidth(0.4) + linetype("4 4") + ffwwqq); draw(circle((0.9392442605110155,5.418992347348019), 5.4997870723801014), linewidth(1.2)); /* dots and labels */ dot((0,0),linewidth(4pt) + dotstyle); label("$O$", (0.21570238228649324,0.4481704999219077), NE * labelscalefactor); dot((-5,0),linewidth(4pt) + dotstyle); label("$A$", (-4.770580274726206,0.4481704999219077), NE * labelscalefactor); dot((5,0),linewidth(4pt) + dotstyle); label("$B$", (5.201985039299192,0.4481704999219077), NE * labelscalefactor); dot((-4,3),linewidth(4pt) + dotstyle); label("$C$", (-3.7962491808271728,3.485790969136537), NE * labelscalefactor); dot((4.776295032025394,1.478852855104097),linewidth(4pt) + dotstyle); label("$D$", (5.030044258022892,1.938323937649839), NE * labelscalefactor); dot((13.30857214203163,0),linewidth(4pt) + dotstyle); label("$M$", (13.512456134320358,0.4481704999219077), NE * labelscalefactor); dot((0.1093794954002178,1.2015494294363533),linewidth(4pt) + dotstyle); label("$K$", (0.33032956980402656,1.651755968856006), NE * labelscalefactor); dot((1.878488521022031,0),linewidth(4pt) + dotstyle); label("$M'$", (2.1070509763257927,0.4481704999219077), NE * labelscalefactor); dot((1.8784885210220088,20.635465563066035),linewidth(4pt) + dotstyle); label("$K'$", (2.1070509763257927,21.08106425307788), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We perform an inversion $\Gamma$ about the semicircle $\Omega$ with center $O$ and diameter $AB$. The points $A, B, C, D$ are fixed under $\Gamma$. Due to preservation of angles under inversion, if $K'$ is the image of $K$ under $\Gamma$ and if $M'$ is the image of $M$ under $\Gamma$, then we need to prove that $\angle OM'K' = 90^\circ$. It can be seen that due to inversion, $K' = AC \cap BD$ and $M' =$ line $\overline{AB} \cap$ circumcircle of $\triangle COD$. Now, we see that $\angle ACB = \angle ADB = 90^\circ$ implies that $AD \perp BK', BC \perp CK'$ and since $O$ is midpoint of segment $AB$, we see that circumcircle of $\triangle COD$ is the nine-point circle of $\triangle K'AB$, now $M' \in$ segment $\overline{AB}$ and $M' \in$ nine point circle of $\triangle K'AB$ and $M'$ is not the midpoint of segment $\overline{AB}$ because $O$ is the midpoint of segment $\overline{AB}$, therefore $M'$ is the foot of the perpendicular from $K'$ to segment $\overline{AB}$ or $\angle K'M'A = 90^\circ = \angle K'M'O$ which is the desired result.

Invert about $\omega$, the semicircle. Note that $A,B,C,D$ were free in the original diagram, and will be sent to themselves post inversion. Next, since $M=AB\cap CD$, we have that $M^* = AB\cap (COD)$. Next, $K=(AOC)\cap (BOD)$, since (AOC) passes through the center of inversion, it will get sent to a line through $AC$. Similarly $(BOD)\to \overline{BD}$. Thus, we have $K^* = AD\cap BC$. Now, simply note that (COD) is the 9-point circle of $\triangle KAB$, thus $M$ must be the foot of the altitude from $K$ so we have $\angle KMO=90$ and we are done $\blacksquare$.

I finally came farther from Kansas Invert at $O$ with diameter $AB$. The inverted diagram is given below. To finish, the purple circle is the nine point circle of $\triangle{KA^{*}B^{*}}$ and the result follows.

Invert around the circle with diameter $\overline{AB},$ and notice that $K^*=\overline{AC}\cap\overline{BD}$ and $M^*=(COD)\cap\overline{AB}.$ Since $(COD)$ is the nine-point circle of $\triangle K^*AB,$ $$\angle MKO=\angle K^*M^*O=90.$$$\square$

Let $AC \cap BD=E$ then its easy to see that $K,E$ are inverses w.r.t. $(O)$ and now taking polars w.r.t. $(O)$ we have that $K \in \mathcal P_E$ but by brokard we have $M \in \mathcal P_E$ thus $\mathcal P_E=MK$ thsu $KM \perp EO$ and since $O,K,E$ are colinear (by the inversion) we have $MK \perp KO$ as desired, thus we are done

Invert around $O$ with $r=AO$. Note that $A$, $B$, $C$ and $D$ stay fixed. Since $OKDB$ is cylic, $K'$, $D'=D$, $B'=B$ are collinear after inversion. Similarly, $K'$, $C'=C$, $A'=A$ are collinear, meaning $K'$ is the intersection of lines $AC$ and $BD$. After inversion $\measuredangle OKM=\measuredangle OM'K'$, so it is enough to show that $\measuredangle OM'K'=90^\circ$. Because $AB$ is the diameter of the semicircle, $\measuredangle ACB=\measuredangle ADB=90^\circ$, so $C$ and $D$ are the foot points of the altitudes on $AK'$ and $BK'$. Since $M$, $D$, $C$ are collinear $OCDM'$ is cyclic. However, $O$ is the midpoint of $AB$, so $(OCD)$ is the nine point circle $(N)$ of $\triangle AK'B$. Hence, $M'=OM \cap(N)\neq O$, meaning $K'M'\perp OM$. Therefore, $\measuredangle OM'K'=90^\circ$.

Invert around $(ABDC)$. Then $(OCD)$ is the nine point circle of $\triangle K^*AB$, so $\angle OKM = \angle OM^*K^* = 90^\circ$, as desired.

Invert the problem around the semicircle. Iran 1996 3rd Exam Problem 2, Inverted wrote: Consider a semicircle of center $O$ and diameter $AB$. Let $C$ and $D$ also lie on this semicircle, with $C$ closer to $A$ than $B$ is. Let $K$ be the intersection of $AC$ and $BD$, and let $M$ be the intersection of $(COD)$ with $AB$. Prove that $KM \perp AB$. Note that by Thales' we have $AD \perp BK$ and $BC \perp AK$, and since $O$ is the midpoint of $AB$, $(COD)$ is the nine-point circle of $ABK$. This immediately implies the desired. $\blacksquare$

Invert about $(AB)$, denoting images with $\bullet'$. We have $K'=\overline{AC} \cap \overline{BD}$, $M'=(CDO) \cap \overline{AB} \neq O$, and we wish to prove that $\angle K'M'O=90^\circ$. Observe that since $O$ is the midpoint of $\overline{AB}$ and $\angle ACB=\angle ADB=90^\circ$, $(COD)$ is the 9-point circle of $\triangle K'AB$, hence $M'$ is the foot of the altitude from $K'$ to $\overline{AB}$, done. $\blacksquare$

Let $CD \cap BD = J$ and $AD \cap BC = L$. Note $K$ is the Miquel point of degenerate quadrilateral $ACBD$. Master Miquel tells us $J$ and $K$ are inverses, and Brocard says $ML$ is the polar of $J$, thus implying the result. $\blacksquare$

Let $\Psi$ be the inversion of arbitrary radius centered at $O$, and for any arbitrary point $P$, let $P'$ denote $\Psi(P)$. Then we first observe: \[\Psi((AOC)) = AC; \phantom{c} \Psi((BOD)) = BD. \]Now $K' = \Psi(K) = AC \cap BD$, and moreover we also procure that $M' = \Psi(M) = (COD) \cap AB$, so that it would suffice to show $\angle K'M'O = 90^{\circ}$. However now observe that as $(COD)$ is the nine point circle of $\triangle K'AB$, we must have $K'M' \perp AB$, as was needed to show. $\blacksquare$