This problem is famous , it is from Russia , in that problem , you have to prove $EF$ is perpendicular to $AB$ and this is just the key step to solve this CWMO problem. Let $AC$ and $BD$ meet at point $L$ , easy to get :$\angle ALB=\frac{\pi}{2}-\angle CAD$ , $\angle CED=\pi-2\angle EDC =\pi-2\angle CAD=2\angle CLD$ , Notice that $EC=DE$ , hence $E$ is the circumcentre of triangle $LCD$ .because $\angle FCL=\angle FDL$ ,hence $L,E,F$ is collinear , Obviously $F$ is the orthocentre of triangle $LAB$ ,hence $EF$ is perpendicular to $AB$ . Join $CM,DM$ , $C,A,M,F$ ; $M,F,D,B$ are both concyclic , $\Longrightarrow$ $\angle EDC=\angle CAF=\angle CMF$ , Hence $E,C,M,D$ are concyclic . QED.

Solution 1 Note that $\angle OCA= \angle OAC = \angle BCE = \alpha$ and $\angle ODB= \angle OBD= \angle ADE = \beta$ Note that since $CE=DE$, $E$ lies on the perpendicular bisector of $CD$. Also, $\angle CFD= \angle FCA + \angle CAF= 90+ \frac{1}{2} \angle COD= 90+ \frac{1}{2} (180- \angle CED)= \newline 180- \frac{1}{2} \angle CED$ Thus $E$ is the circumcenter of $\triangle CFD$, so $\angle EFC= \angle ECF= \alpha$ and thus $\angle CEF= \angle COA$. Thus $CEOM$ is cyclic, similarly $DEMO$ is cyclic, so $CEDOM$ is cyclic. In fact, here is a more interesting and quicker solution: Solution 2 Rotate $\triangle DOB$ about $O$ until $B$ coincides with $A$, let $D$ rotate to $D'$. Note that we have $CO=OD'$ and $CE=ED$. Also, $\angle COD'=180- \angle COD =\angle CED$. Next, $\angle ACO=\angle CAO= \angle ECF$ and $\angle AD'O= \angle ODB= \angle OBD= \angle EDF$. Thus, $ECFD \sim OCAD'$. It follows that $\angle CEF= \angle COA$, so $CEOM$ is cyclic, similarly $DEMO$ is cyclic, so $CEDOM$ is cyclic.

Just apply Pascal theorem to $ACCBDD$

mahanmath wrote: Just apply Pascal theorem to $ACCBDD$ Brilliant idea! I actually went to this year's CWMO, and to be honest, almost the entire Hong Kong team used coordinate geometry to solve both geometry problems - i guess that's our style.

Andy Loo wrote: mahanmath wrote: Just apply Pascal theorem to $ACCBDD$ Brilliant idea! I actually went to this year's CWMO, and to be honest, almost the entire Hong Kong team used coordinate geometry to solve both geometry problems - i guess that's our style. Wow, that's quite amazing. I also used Pascal to solve this. This is my friend's solution: $\angle CFD=\frac12(\widehat{CD}+\widehat{AB})=\frac12\widehat{CD}+90^{\circ}$ and $\angle CED_{maj}=360^{\circ}-\frac12(\widehat{CD}_{maj}-\widehat{CD})=360^{\circ}-\frac12(360^{\circ}-2\widehat{CD})=180^{\circ}+\widehat{CD}$. Thus $\angle CED_{maj}=2\angle CFD$, therefore $C,F,D$ lie on a circle centered at $E$, which gives $EC=EF=ED$. So $\angle BFM=\angle CFE=\angle FCE=\frac12\widehat{BC}=90^{\circ}-\frac12\widehat{AC}=90^{\circ}-\angle FBM$, thus $\angle FMB=90^{\circ}$. So $\angle EMO=\angle EDO=\angle ECO=90^{\circ}$, which implies that $E,D,M,O,C$ are cyclic.

Dear mathlinkers, yes, the idea of using Pascal is good in order to have a synthetic proof. After, we can can show thet this circle goes through the midpoint of AB... Sincerely Jean-Louis

Let $CD$ cut $AB$ at $K$, $AC$ cut $BD$ at $G$. $GF$ is $k$, the polar of $K$ $CD$ is $e$, the polar of $E$. As $K$ is on $e$, $E$ must be on $k$, the polar of $K$. Thus $G, E, F$ are all on $k$, they are collinear and $GM \perp KO$. It follows that $E, C, M, O, D$ are concyclic.

Attachments:

From figure above, $\angle AFC=\angle AGB\implies \angle AGB=\frac{\angle COA+\angle BOD}{2}$ $=90^\circ-\frac{\angle COD}{2}=\angle CDO=\angle DCO$, so $CO, DO$ are tangent to the circle $(CDG)$, hence $E$ is its circumcenter, $\iff GE\bot AB$. Best regards, sunken rock

There's another simple solution. Suppose $AC$ and $BD$ meet at $G$ (still using the figure above), we easliy get $F$ is the orthocenter of $\triangle ABG$. Let $E'$ be the midpoint of $GF$, then we have $\angle GCE'=\angle CGE'=\angle ABF=\angle BCO$, hence $\angle E'CO=\angle GCF=90^\circ$, which means $E'C$ is tangent to circle $O$ at $C$. Similarly, we have $E'D$ is tangent to circle $O$ at $D$, so $E'$ is $E$, thus $EM\bot AB$. Finally, we have $E, C, M, D$ are on the nine-point-circle of $\triangle ABG$.

Solution: Extend $CD$ to meet $AB$ at $X$.Note that $CD$ is the polar of $E$ and it passes through $X$.So the polar of $X$ must pass through $E$.Also it is well known that the polar of $X$ passes through $F$.So $EF$ is the polar of $X$ which means that $EF \perp OX \implies EF \perp AB$.Now the rest is easy:Note that $MFDB$ concyclic(why?) which means $\angle{ECD}=\angle{CBD}=\angle{FBD}=\angle{FMD}=\angle{EMD}$ and everything is oK.

Just a question, do u still need to deal with the config issue where F could be inside or outside the semicircle, or it's ok to just deal with the inside case?

Let $AB \cap CD=P$ and $AC \cap DB=Q$. Then by Brocard's theorem $QF$ is the polar of $P$ , and hence $QF \perp AB$. Now $CD$ is the polar of $E \implies P$ lies on the polar of $E$. By La hire's theorem the polar of $P$ which is $QF$ passes through $E$. So $Q,E,F,M$ collinear. Hence $\angle EMA =90$. So $\angle CME=\angle CAD=\angle ECD \implies C,E,D,M$ is concyclic.

We claim that $E$ is the circumcenter of triangle $CDF$, indeed \[\angle DEC=2 \angle DFC \iff 180^{\circ}-\angle EDC-\angle ECD=360^{\circ}-2\angle A-2\angle B\iff \angle A+\angle B-\angle CBD=\angle A+\angle DBA=90^{\circ}\]which is true as $AB$ is the diameter of the circle with center $O$. So, we now prove that quadrilateral $OCEM$ is cyclic, \[\angle COM=2\angle B=2\angle CDF=\angle CEF \implies \textrm{OCEM is cyclic. $\square$}\]Then, we show that quadrilateral $OCED$ is cyclic, \[\angle CED=180^{\circ}-\angle EDC-\angle ECD=180^{\circ}-2\angle CBD=180^{\circ}-\angle COD \implies \textrm{OCED is cyclic. $\square$} \]Therefore, since quadrilaterals $OCEM$ and $OCED$ are cyclic, we have quadrilateral $ECMD$ is cyclic. $\blacksquare$