ISL 2002 G8

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Comment:This problem seemed to be relatively easy for a G8.

Let $S_{1},S_{2}$ be circles meeting at the points $A,B$. A line through $A$ meets $S_{1}$ at $C$ and $S_{2}$ at $D$. Points $M,N,K$ lie on the line segments $CD,BC,BD$ respectively,with $MN$ parallel to $BD$ and $MK$ parallel to $BC$. Let $E$ & $F$ be points on those arcs $BC$ of $S_{1}$ and $BD$ of $S_{2}$ respectively that do not contain $A$. Given that $EN$ is perpendicular to $BC$ and $FK$ is perpendicular to $BD$,prove that \[\angle EMF=90^{\circ}\].

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I can't check myself from giving the solution

Let $G$ be the other point of intersection $FK$ with the arc $BAD$. Since $\frac{BN}{NC}=\frac{DK}{KB}$ and $\angle CEB=\angle BGD$,we have $\triangle CEB\sim \triangle BGD$. Now we have $\frac{BN}{NE}=\frac{DK}{KG}=\frac{FK}{KB}$. So from $BN=KM,BK=MN$,we deduce that $\frac{MN}{NE}=\frac{FK}{KM}$. But we also have $\angle MNE=\angle MNB+90^{\circ}=\angle MKB+90^{\circ}=\angle MKF$. Thus $\triangle MNE\sim \triangle FKM$. Finally,
$\angle EMF=\angle NMK-\angle NME-\angle KMF=\angle NMK-\angle NME-\angle NEM=
\angle NMK-90^{\circ}+\angle BNM
=90^{\circ}$