Very nice problem. Let $LD$ cut $(AOC)$ at $G$, $LC$ cut $(BOD)$ at $F$, easy that $GA\parallel BD$ and $BF\parallel AC$, so both $\Delta GAD$ and $\Delta CBF$ are right-angled. Let $E\equiv AD\cap BC$. As $\angle CFB = 180^{\circ} -\angle LDB = \angle AGD$, $\Delta GAD\sim \Delta CBF$ and thus $CLED$ is cyclic. As $OD = OB$, $\angle OLB = \angle GLO$, as $OA = OC$, $\angle ALO = \angle OLF$, thus $\angle ALB = \angle ALD$ and it follows that $\angle LAD = \angle LBC$ or $LEBA$ cyclic. Notice that $LE$ is the radical axis of $(LEBA)$ and $(LEDC)$, but as $MC. MD = MA. MB$, then $M, L, E$ collinear. On the other side, easy to see that $AC, LO, DB$ concur at a point, say $P$, hence $LM$ is the polar of $P$ and $LM\perp OP$.

Under inversion about the semi-circle, $\odot AOC\mapsto AC$ and $\odot BOD\mapsto BD$, so $K\mapsto K':=AC\cap BD$. Obviously, O,K,K' are collinear. Obviously, $K$ lies on the polar of $K'$. By self-polarity of diagonal triangle of cyclic quad $ABCD$, we have that $M$ lies on the polar of $K'$, i.e., $MK\perp KO$.

Dear Mathlinkers, see http://www.artofproblemsolving.com/Forum/viewtopic.php?f=150&t=353558 Sincerely Jean-Louis

what the purpose of this link?

Dear Mathlinkers, now the correct link http://www.artofproblemsolving.com/Forum/viewtopic.php?t=4919 Sincerely Jean-Louis

On the other side, easy to see that $ AC, LO, BD $ concur at a point, say $ P $, hence $ LM $ is the polar of $ P $ and $ LM\perp OP $. I cannot understand what $ LM $ is the polar of $ P $ means, can someone explain it for me in detail? also , "Under inversion about the semi-circle", what inversion means?

*Suppose L = K (sorry) AC, BD, OK are concurrent at Q as they are radical axises of three circles. Let AD intersect BC at P. Let R be the Miquel point of quad ABCD. So R lies on QM and OR is perpendicular to QM. As DBMR is cyclic, QR*QM=QD*QB Again, as KDBO is cyclic, QK*QO=QD*QB So, QR*QM=QK*QO. Ss OKRM is cyclic and <OKM = <ORM = 90°.

Note*: In my solution $K$ represents $L$. Let $\overline{AD} \cap \overline{BC}=R$, $\overleftrightarrow{AC} \cap \overleftrightarrow{BD}=Z$ and $\overleftrightarrow{MR} \cap \overleftrightarrow{ZO}=K^* $. We know that $\triangle MRZ$ is self-polar triangle and $O$ is its orthocenter. In other word $\overline{ZO} \perp \overline{MK}^*$. So we would be done if we can prove that $K^* \equiv K$ or $(AOC) \cap (BOD)= O, K^*$. In other word, we've to show that quadrilateral $ACK^*O$ and quadrilateral $BDK^*O$ are cyclic. Let $U$ be the miquel point of cyclic quadrilateral $ABDC$. And $\overleftrightarrow{ZR} \cap \overline{AM}=E$. As $O$ the orthocenter of $\triangle MRZ$, $\angle ZK^*M=\angle ZEM= \angle RUM= 90^o$ [it is well-known that $O, R, U$ are collinear.] So, quadrilateral $AMQC, UMER, REOK^*$ are cyclic. Now we'll use power point. $\overline{ZC}\times \overline{ZA}=\overline{ZU}\times \overline{ZM}=\overline{ZR}\times \overline{ZE}=\overline{ZK}^*\times \overline{ZO}$ so $\overline{ZC}\times \overline{ZA}=\overline{ZK}^* \times \overline{ZO}= \overline{ZD}\times \overline{ZB}$. Which concludes that quadrilateral $ACK^*O$ and quadrilateral $BDK^*O$ are cyclic. So, $K^*\equiv K$ $Q.E.D$

Super-quick by Inversion. This is Iran 1996 as well. micliva wrote: Let be given a semicircle with diameter $AB$ and center $O$, and a line intersecting the semicircle at $C$ and $D$ and the line $AB$ at $M$ ($MB < MA$, $MD < MC$). The circumcircles of the triangles $AOC$ and $DOB$ meet again at $L$. Prove that $\angle MKO$ is right. L. Kuptsov Invert around the semicircle! Let $X^*$ denote image of $X$ under inversion.

Quote: Let be given a semicircle with diameter $AB$ and center $O$, and a line intersecting the semicircle at $C$ and $D$ and the line $AB$ at $M$ ($MB < MA$, $MD < MC$). The circumcircles of the triangles $AOC$ and $DOB$ meet again at $K$. Prove that $\angle MKO$ is right Solution: Let $AC \cap BD=E$ and $AD \cap BC=F$. Obviously, $EF \cap AB$ and $OK$ passes through $E$. Let $EF \cap BC=P$, then, $$EF ~ \cdot ~ EP = ED ~ \cdot ~ EB = EK ~ \cdot ~ EO \implies KOPF \text{ cyclic }$$Hence, $K$ is the $E-$humpty point in $\Delta AEB$ $\implies$ $K$ $\in $ $\odot (ECD)$, $\odot (AFB)$. By Radical Axes Theorem on $\odot (ECD)$ $, \odot (ACDB)$ $, \odot (AFB)$ $\implies$ $K-F-M$ $\implies$ $\angle MKO=90^{\circ}$

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)

micliva wrote: Let be given a semicircle with diameter $AB$ and center $O$, and a line intersecting the semicircle at $C$ and $D$ and the line $AB$ at $M$ ($MB < MA$, $MD < MC$). The circumcircles of the triangles $AOC$ and $DOB$ meet again at $L$. Prove that $\angle MKO$ is right. L. Kuptsov

By radical axis theorem on $\odot AOC,\odot BOD,\odot ABC$ we get $AC,BD,OK$ concur at a point ,say $E$. Next,applying Miquel's three circle theorem on $EAB$ and $O\in AB,D\in BE,C\in AE$ we get $K\in\odot ECD$.Let $H$ be the orthocentre of $EAB$. Since,$\odot ECD$ is the circle with diametre $EH$,and $EO$ is $E$-median of $EAB$,hence $K$ is the $E$- HM point of $EAB$. On the other hand,$M$ is the $E$-Ex point of $EAB$.Hence $MK\perp EO$.

We claim that $AMKB$ must be a cyclic quadrilateral. Our proof is that $$\angle AMB = 180^{\circ} - \angle BAM - \angle MBA = 180^{\circ} - \angle BAD - \angle CBA = 180^{\circ} - \frac{\angle COA + \angle BOD}{2}$$and since $\angle COA = 180^{\circ} - \angle OAC - \angle ACO = 180^{\circ} - 2\angle ACO$ and $\angle BOD = 180^{\circ} - \angle ODB - \angle DBO = 180^{\circ} - 2\angle ODB$, then plugging into our equation gets $$180^{\circ} - \frac{\angle COA + \angle BOD}{2} = 180^{\circ} - \frac{360^{\circ} - 2\angle ACO - 2\angle ODB}{2}$$$$ = \angle ACO + \angle ODB = \angle AKO + \angle OKB = \angle AKB.$$Now, we see that $\angle MKO = \angle MKB - \angle OKB$. Since $AMKB$ is a cyclic quadrilateral, then $\angle MKB = 180^{\circ}-\angle BAM$ and $\angle BAM = \angle BAD$ gets that $\angle MKO = 180^{\circ} - \angle BAD - \angle OKB$ and $\angle OKB = \angle ODB = \angle DBO$. Now, we know that $\angle BAD = \angle DBA = 90^{\circ}$ which means that $$\angle MKO = 180^{\circ} - \angle BAD - \angle DBO = 180^{\circ} - \angle BAD - \angle DBA = 180^{\circ} - 90^{\circ} = 90^{\circ}$$and we are now done because $\angle MKO =\angle OKM = 90^{\circ}$.

$K$ is the Miquel Point of $ADCB$ by definition, hence done by Master Miquel Theorem. Alternatively, $\overline{CD}$ inverts upon the circumcircle to the nine-point circle of triangle $ABK^*$, so the rest follows immediately.

By the radical center theorem $T=AC\cap BD\in OL,$ and by the Miquel theorem $L\in \odot (CTD).$ Since $TL$ bisects $AB,$ it's a symmedian in $\triangle TCD,$ so $OC$ tangents to $\odot (TDC),$ i.e. $T,L$ are inverse wrt $\odot (ABCD).$ Hence $ML$ is the polar of $T$ wrt this circle and conclusion follows.

We invert about the circle with center $O$ and radius $AO$. This means that all of $A$, $B$, $C$, and $D$ do not move. However, $K^*$ is the intersection of $AC$ and $BD$, and $M^*$ is the point on $AB$ such that $CDOM^*$ is concyclic. It remains to prove that $OM^*K^* = 90$. Specifically we claim that $CDM^*O$ is the nine-point circle. This is evident since $\angle ACB = 90 = \angle ADB$ and $AO = OB$. $\blacksquare$

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. $\blacksquare$

Invert with respect to $(ACBD)$ then it suffices to show the foot of the altitude $M'$ from $L'=AC \cap BD$ to $AB$ lies on $(OCD)$, however this is trivial as $(OCD)$ is the nine-point circle of $(ABL')$ so $M'$ lies on it and we can conclude.

We invert the diagram about $\omega$. By how polars are defined, $M'$ lies on the polar of $M$ with respect to $\omega$. $K'=AC\cap BD$, and thus by Brocard's Theorem, $K'$ also lies on the polar of $M$ with respect to $\omega$, and thus $OM \perp M'K'$. Thus we have $\angle MKO = \angle K'M'O = 90^{\circ}$, as desired.

$ $

Invert around the circle, and define $AD\cap BC=Y$ and the Miquel point as $Z$. Then $Y$ and $Z$ swap and $K$ and $X$ swap. Since $\angle OZM=90^{\circ}$, it suffices to show that $MZKO$ is cyclic, or, after inverting, that $M'YX$ are collinear. But $XY$ is the pole of $M$, done. $\blacksquare$

We can restate the problem as follows: In triangle $\Delta ABC$, $E,F$ are feet of $B,C$ altitudes. $M$ is midpoint of $BC$ then prove that $(BME) \cap (CMF)$ is the $A$ humpty point. Now inversion about $M$ with radius $MB$ yields the desired.

[asy][asy] size(8cm); pair A = dir(180); pair B = dir(0); pair O = 0.5*(A+B); pair C = dir(145); pair D = dir(95); pair K = intersectionpoints(circumcircle(A, O, C), circumcircle(B, O, D))[0]; pair M = extension(A, B, C, D); draw(arc(midpoint(A--B), B, A)); draw(M--B^^M--D^^A--C^^B--D); draw(circumcircle(A, O, C)); draw(circumcircle(B, O, D)); draw(M--K--O, red+linewidth(1)); pair S = M-K; pair T = O-K; draw(K+dir(S)*0.08--K+dir(S)*0.08+dir(T)*0.08--K+dir(T)*0.08, red); dot("$A$", A, dir(225)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$K$", K, dir(K)); dot("$M$", M, dir(M)); dot("$O$", O, dir(-90)); [/asy][/asy] Now, inverting about the semicircle, the image becomes [asy][asy] import geometry; size(6cm); pair A = dir(180); pair B = dir(0); pair O = 0.5*(A+B); pair C = dir(145); pair D = dir(95); pair K = intersectionpoint(line(A, C), line(B, D)); pair M = intersectionpoints(line(A, B), circumcircle(O, C, D))[0]; draw(arc(midpoint(A--B), B, A)); draw(A--K--B--cycle^^C--D); draw(circumcircle(C, O, D)); draw(K--M--O, red+linewidth(1)); pair S = K-M; pair T = O-M; draw(M+dir(S)*0.08--M+dir(S)*0.08+dir(T)*0.08--M+dir(T)*0.08, red); dot("$A$", A, dir(225)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$K^{*}$", K, dir(K)); dot("$M^{*}$", M, dir(-90)); dot("$O$", O, dir(-90)); [/asy][/asy] Since $\overline{BC}\perp\overline{AK^{*}}$, $\overline{AD}\perp\overline{BK^{*}}$, and $O$ is the midpoint of $\overline{AB}$, $(COD)$ is the 9-point circle of $\Delta ABK^{*}$. Hence, we see that $\angle OM^{*}K^{*}=90^\circ$. Lastly, \[ 90^\circ = \angle OM^{*}K^{*} = \angle OKM \]$\blacksquare$