Since $\ell$ is antiparallel to $DC$ WRT $EC,ED,$ it follows that $EH$ is the E-circumdiameter of $\triangle EDC.$ Let $M$ be the orthogonal projection of $E$ on $HK$ and $EM$ cuts $DC$ at $T.$ From the cyclic quadrilateral $EKHG$ (due to its right angles <EKG and <EHG) we deduce that $\angle HEG=\angle HKG=\angle MKT=\angle KET.$ Since $EK,EH$ are isogonals WRT $\angle DEC,$ then $EM,EG$ are also isogonals WRT $\angle DEC.$ From $\triangle DEC \sim \triangle AEB,$ we have then $\angle FEB=\angle GEC=\angle MED$ $\Longrightarrow$ $F,E,M$ are collinear, i.e. $EF \perp HK.$

Here is my solution to this nice problem. Let $EH$ meets $AB$ at $P$. Then we have to prove that $\angle{PEF}+\angle{EHK}=90$. But we have that $\angle{EHK}=\angle{EGK}=90-\angle{KEG}$. So we have to prove that $\angle{KEG}=\angle{FEP}$. But $DEC$ and $AEB$ are similar and the angles $KEG,FEP$ are these that are formed by the median and the height to these similar triangles so they are equal and we are done.

Let ${U}=HK \cap EF.$ The condition can be written as $\widehat{GKH}=\widehat{KEU}$,but $G,H,K,E$ are cocyclic, so $\widehat{GKH}=\widehat{GEH}$.Denote ${\widehat{GEH}=x,\widehat{HGE}=90-x}$, denote $P=GE \cap AB$ so $\widehat{EPF}=90-x$ Denote $R=EK \cap AB$. We have to prove that $\widehat{REF}=x$ Denote $M=EF \cap DC$, so we have to prove that $\widehat{EMG}=\widehat{EPF}$, but we have ${\widehat{PEF}=\widehat{MEG}}$, so we have to prove that $\widehat{DGE}=\widehat{EFA}$, but this is well-known.(prove that $\widehat{AEF}=\widehat{DEG}$(use $\frac{AE}{EB}=\frac{sin \widehat{FEB}}{sin \widehat{AEF}}$.

Let $EF\cap l=X$. Then $\angle FXH=\angle AFE$. We claim $\angle AFE=\angle EGD$. Indeed, $\triangle AEB\sim \triangle DEC$, and as $F$ is the midpoint of $AB$, $G$ the midpoint of $CD$, we must have $\triangle AFE \sim \triangle DGE$. Now let $EX\cap KH=Y.$ THen $EKGH$ is cyclic. But $\angle EKH=\angle EGH$ and $\angle KEX=\angle GEH$, implying $\angle EYK=90^o$ as desired.

Does anyone know anything about the results? Thank you!

The medal cut-offs are known : 10 for bronze, 30 for gold, but I'm not sure if 17 or 21 are required for silver. Some of the countries posted their results, but the complete official results can't be seen on http://www.bmo2011.lbi.ro yet. I don't know what's wrong with the page.

Simson's line (E,KGH) (suppose $EX'$ perpendicular to $HK$) + Proof $\angle{BEF}=\angle{X'ED}$ by Similarity Done

matrix41 wrote: Simson's line (E,KGH) (suppose $EX'$ perpendicular to $HK$) + Proof $\angle{BEF}=\angle{X'ED}$ by Similarity Done Well Done ! I had also notice it but my proof was little different : to prove that KH is the Simpson Line of TriangleKGH it is enough to prove that the steiner line of this triangle is parallel to simpson line. This can be easily done if we recall the Brocard Lemma to help us, saying : ''The quadrilateral ABCD is inscribed in the circle k with center O. Let E = AB∩CD, F = AD∩BC, G = AC∩BD. Then O is the orthocenter of the triangle EFG. ''

oh yes!i proof some lemma! lemma1:$\angle DEK = \angle HEC$ suppose $EH$ intersect $AB$ on $P$ because $l$ is parallel to $AB$ $\angle BPE =90$ so $\angle KEC = \angle PEB= \angle DEH$ so $\angle DEK = \angle HEC $! end of proof of lemma 1. suppose $EF$ intersect $HK$ at $Q$...! lemma2:$EQ$ is isogonal of $EG$ wrt $DE$ and $EC$ i proof that $\frac {\sin CEG}{\sin GED} = \frac{\sin DEQ}{\sin CEQ} $ proof of lemma 2: $\frac {\sin CEG}{\sin GED} = \frac{ED}{EC}$ we know that $\angle DEQ =\angle FEB$ and $\angle DEC=\angle AEB$ $\frac {\sin BEF}{\sin FEA} = \frac {EA}{EB}$ know we must to proof $\frac {EA}{EB}= \frac{ED}{EC}$ and it is so easy that we can proof by $\triangle AEB \sim \triangle DEC$ end of proof of lemma 2. know from lemma1 and lemma2 $\rightarrow \angle KEQ = \angle GEH $ so $EQ$ is isogonal of $EG$ wrt $EK$ and $EH$ know i need lemma 3 lemma 3:theoream 2 from isogonal conjugation with respect to a traingle frome Darij grinberg from lemma 3 and lemma 2 and lemaa 1 $\rightarrow$ problem proof! lemma 3:http://www.cip.ifi.lmu.de/~grinberg/Isogonal.zip

easy angle chasing leads to $ EF $ bring the symmedian of triangle $ EDC $ . Now some more easy angle chasing leads to the desired result.

My solution is probably the same as waver123's.

My Solution: Let's say that $\angle CAB=\angle CDE=\alpha, \ \angle ABD=\angle ACD=\beta , \ \angle DEG=\phi$ and $\angle AEF=\lambda$ In the triangle $AEB$ we have: \[\frac{sin \ \alpha}{sin \ \beta}=\frac{sin \ \lambda}{sin \ (\alpha+\beta+\lambda)} \ (\star)\]and similarly in triangle $DEC$ we have \[\frac{sin \ \alpha}{sin \ \beta}=\frac{sin \ \phi}{sin \ (\alpha +\beta + \phi)} \ (\star\star)\]Combinig $(\star)$ and $(\star\star)$ we get \[\frac{sin \ \phi}{sin \ (\alpha+\beta+\phi)} = \frac{sin \ \lambda}{sin \ (\alpha +\beta + \lambda)}\]Solving the equation we get $\alpha = -\beta$ (Which is impossible) then $\phi=\lambda$ By angle Chasing we get $\angle EKT=\beta+\phi$ Where $T=HK\cap FE$ and since $\phi=\lambda$ we get $\angle KET =90^{\circ}-\beta - \phi$ which follows $\angle ETK=90^{\circ}$

let $l \cap AC=J$ and let $HK \cap EF=M$ we must prove that $\angle{MEK}=90-\angle{EKM}=90-\angle{EGH}$ $\angle{EGH}=\angle{GEJ}+\angle{GJE}$ $\angle{GJE}=\angle{EAB}=\angle{EDK}$ $\angle{GEJ}=\angle{FEB}=\angle{DEM}$ we know that $\angle{EDK}+\angle{DEM}=90-\angle{EKM}$ so we are done

Let $l \cap FE=J$ and $l \cap AC=L.$ Since $AB \parallel l,$ we have $\angle AFE=\angle EJH.$ Then from $\triangle AFE \sim \triangle DGE$ and cyclic quadrilateral $EKHG$ we have $\angle AFE=\angle DGE=\angle KHE=\angle LHE.$ So, $\angle EJH= \angle LHE.$ It means that $\triangle EJH \sim \triangle EHL.$ Since $\angle EHJ=90^\circ,$ we get $\angle ELH=90^\circ,$ which means that $EF \perp HK.$

My solution. let $Y$ be the intersection of $AB$ with $HE$ and let $X$ be the intersection of $FE$ with $HK.$ since the quadrilateral $ABCD$ is Cyclic we have, $\measuredangle GDE=\measuredangle CDB=\measuredangle CAB=\measuredangle EAF.$ and, $\frac{AE}{DE}=\frac{AB}{DC}=\frac{\frac{1}{2}AB}{\frac{1}{2}DC}=\frac{AF}{DG}$ (since, the triangles $ABE$ and $EDC$ are simillar). hence, the triangles $DEG$ and $AFE$ are simillar and, $\measuredangle EGD=\measuredangle AFE.$ now, we see that the quadrilateral $EGHK$ is Cyclic since, $\measuredangle GHE=\measuredangle GKE=90^\circ.$ therefore, $\measuredangle AFX=\measuredangle AFE=\measuredangle EGD=\measuredangle EGK=\measuredangle EHK=\measuredangle EHX.$ and thus, the quadrilateral $YFHX$ is Cyclic. hence, $\measuredangle HXF=\measuredangle HYF=90^\circ.$ since, $\ell$ and $AB$ are parallel and $EH$ is perpendicular to $\ell.$

manifestdestiny wrote: Let $EF\cap l=X$. Then $\angle FXH=\angle AFE$. We claim $\angle AFE=\angle EGD$. Indeed, $\triangle AEB\sim \triangle DEC$, and as $F$ is the midpoint of $AB$, $G$ the midpoint of $CD$, we must have $\triangle AFE \sim \triangle DGE$. Now let $EX\cap KH=Y.$ THen $EKGH$ is cyclic. But $\angle EKH=\angle EGH$ and $\angle KEX=\angle GEH$, implying $\angle EYK=90^o$ as desired. how is AFE similar to DGE

$EF$ is the symmedian of triangle $ECD$. So $EF$ and $EG$ should be isogonal, some angle chase finishes the problem.