mathisreal wrote: Let $AB$ be a diameter of a circle $\Gamma$ with center $O$. Let $CD$ be a chord where $CD$ is perpendicular to $AB$, and $E$ is the midpoint of $CO$. The line $AE$ cuts $\Gamma$ in the point $F$, the segment $BC$ cuts $AF$ and $DF$ in $M$ and $N$, respectively. The circumcircle of $DMN$ intersects $\Gamma$ in the point $K$. Prove that $KM=MB$. [Still looking for euclidean solution ] First, use Menelaus' theorem on $\triangle BCO$ and transversal line $MEA$. We have $\frac{BM}{MC} \cdot \frac{CE}{EO} \cdot \frac{OA}{AB} = -1 \Rightarrow BM=2MC$. Also consider that $\angle ECN=\angle ABC=\angle EFN \Rightarrow CENF$ is a cyclic quad. Hence $\angle BAM=\angle NCF=\angle NEM \Rightarrow EN\parallel AB$. Since $OE=EC$, then $BN=NC$. Let $\Gamma$ be the unit circle and $A=-1, B=1, C=c$ in a complex plane. We have $D=\frac{1}{c}$. Since $N=BC\cap DF$, then by computation, \[n=\frac{df(b+c)-bc(d+f)}{df-bc} \Rightarrow f=\frac{c(2+c)}{1+2c}.\] Let $K'$ be the point on $\Gamma$ such that $\stackrel\frown{FC} = \stackrel\frown{CK'}$. Then $fk'=c^2 \Rightarrow k'=\frac{c(1+2c)}{2+c}$. Note that $BN=NC$ and $BM=2MC$. Then $n=\frac{1+c}{2}$, $m=\frac{1+2c}{3}$. One can check that \[\frac{n-m}{k-m}:\frac{n-d}{k-d} \in \mathbb{R}.\]$DMNK'$ is a cyclic quad. Then $K'=K.$. Also, $\stackrel\frown{FC} = \stackrel\frown{CK}$. Consider that $DMNK$ is a cyclic quad gives $\angle DKM=\angle DNB= \stackrel\frown{BD}+\stackrel\frown{FC}$. Then \[\angle BKM=\angle DKM-\angle DKB= \stackrel\frown{FC}=\angle FDC=\angle CDK=\angle MBK.\] $\therefore BM=MK$ as desired.

an Euclidean Solution?

Let $G$ be on segment $DB$ such that $BG=BD/3$ and let $I$ be $(OGM) \cap (ABC)$. Since $\triangle COA \sim \triangle CBD$, $NC=NB \implies EN = AB/4 \implies CM=BG$. So $DGMN$ is cyclic. The spiral similarity centered at $O$ that sends $DB$ to $BC$ also sends $G$ to $M$, then $\triangle OGM \sim \triangle OBC \implies OGBIM$ is cyclic, which implies $GM \parallel BI$. But $KD, GM, BI$ are concurrent by radical axis, then $KD \parallel GM \implies KM=DG=BM.$

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My solution Let $(DMN)$ intersects $DB$ at $P \neq B$ so $DP=2PB$. It is easy to show that $OM=OP$ ,i.e $MP \parallel KD$, it implies $KMPD$ is trapezoid So we have $KM=DP=BM$, as desired. Q.E.D

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