Dear Mathlinkers, the perpendicular to OM through which point? Sincerely Jean-Louis

Let the line through $P$ parallel to $CD$ intersect $AB$ at $N$.$AB$ is the polar of $P$ and $M$ lies on $AB$.So by La Hire's theorem the polar of $M$ passes through $P$.Again $OM\perp NP$.So $NP$ is the polar of $M$.Again $MN$ is the polar of $P$.So $MP$ is the polar of $N$.This implies $N$ and $M$ divides $AB$ harmonically.Now we consider the harmonic pencil $PN,PA,PM,PB$.As $PN$ and $CD$ are prallel,they intersect at infinite.But this point at infinite,$C,M$ and $D$ is harmonic.So $M$ is the midpoint of $CD$.

Firstly note that $PA=PB$. So $\angle PAB=\angle PBA$. Now join $O,A;O,B;O,C;O,D$. Then $\angle OMC=\angle OAC=90^{\circ}$ implies that $O,C,A,M$ are concyclic. Similarly $O,M,D,B$ are concyclic. Now, $\angle COM=\angle PAB=\angle PBA=\angle DBM=\angle DOM$. Also $OM\perp CD$. Thus by the ASA form of congruence, $\triangle COM \cong \triangle DOM$ and finally $CM=DM$.

First, we denote the mid-point of $ A,B $ is $ N $ By easy angle chasing, we have $ O,M,A,C $ and $ O,M,B,D $are concyclic. Therefore, $ \triangle COM\sim\triangle AON $ and also $ \triangle DOM\sim\triangle BON $ $ \Rightarrow \triangle COM\cong\triangle DOM $ $ \Rightarrow CM=DM $

Olemissmath wrote: From the point $P$ outside a circle $\omega$ with center $O$ draw the tangents $PA$ and $PB$ where $A$ and $B$ belong to $\omega$.In a random point $M$ in the chord $AB$ we draw the perpendicular to $OM$, which intersects $PA$ and $PB$ in $C$ and $D$. Prove that $M$ is the midpoint $CD$.

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The problem is easy. We see that $O,\,M,\,A,\,C$ are concyclic, therefore $\angle MCO=\angle MAO.$ Similarly, we also have $\angle MDO =\angle MBO.$ On the other hand, the triangle $AOB$ is isosceles at $O,$ so $\angle MAO=\angle MBO.$ It follows that $\angle MCO=\angle MDO,$ or $\triangle COD$ is isosceles at $O,$ from which we deduce the result.

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Dear Mathlinkers, this nice problem has a relation with the butterfy theorem... See for example http://perso.orange.fr/jl.ayme vol. 7, A new metamorphosis of the butterfly theorem..., p. 42-43. Sincerely Jean-Louis

I have another quick solution. The perpendicular from $ O $ to $ PC $ is $ A $, from $ O $ to $ PD $ is $ B $, and from $ O $ to $ CD $ is $ M $, and since $ A, B, M $ are collinear, by Simpson, $ PCOD $ is cyclic. Therefore, $ \angle OCD=\angle OPD=\angle OPB=\angle OPA=\angle OPC=\angle ODC $, so $ OCD $ is isosceles and $ CM=MD $.

WLOG, suppose $AM\le MB$. Just remark that $O$ is the Miquel point of the (complete) quadrilateral $AMDPBC$, hence $PCOD$ is cyclic. As $PO$ bisects $\widehat{CPD}$, it follows that $OC=OD$, i.e. $M$ is the midpoint of $CD$.

Lolwut how is this a TST #3? Observe that quadrilaterals $MOAC$ and $MDOB$ are cyclic since $OM \perp CD, OB \perp DB,$ and $OC \perp AC.$ Then, $\angle MAO = \angle MCO$ and $\angle MBO= \angle MDO.$ Since $OA=OB,$ $\angle OBM = \angle OAM,$ so $\angle ODM = \angle OCM \implies \triangle ODM \cong \triangle OCM \implies MC= MD.$ $\blacksquare$

Very Easy! $\measuredangle OAC = \measuredangle OMC = \frac{\pi}{2} \implies M \in \odot AOC$. Similarly, $M \in \odot BOD$. Now in right angled $\triangle BOD$ and $\triangle AOD$, $OA=OB$ and $\angle AOC = \angle AMC = \angle BMD = \angle BOD$. $\therefore \triangle AOC \cong \triangle BOD \implies OC=OD$ and as $OM \perp CD$, we must have $M$ is the the midpoint of $CD$.

Let the line through $M$ perpendicular to $OM$ meet $\omega$ at $X$ and $Y.$ Then $M$ is the midpoint of $\overline{XY}$ on account of $OM \perp XY.$ Then (a degenerate case of) the Butterfly Theorem applied to chords $AB$ and $AB$ of $\omega$ implies that $M$ is the midpoint of $AA \cap XY$ and $BB \cap XY$, as desired. $\square$

I rename $M$ to $C$, and $C$ and $D$ to $D$ and $E$. Let line $CD$ hit the circumcircle at $F$ and $G$. We claim circles $(OFG)$ and $(ODE)$ are tangent; invert about the circumcircle, and denote inverses with a $'$. $D'$ and $E'$ clearly lie on $(OFG)$, which means the circles are homothetic, since $O, D, D'$ are collinear. The two circles invert to parallel lines $D'E'$ and $FG$, which means that $\angle ODE=\angle OE'D'=\angle OED$ as desired.

Let $Q$ be the intersection between $AB$ and the line through $P$ parallel to $CD$. Since $M\in\text{polar}(P)$ by La Hires $P\in\text{polar}(M)$, and since $CD\perp OM\implies PQ\perp OM$, $PQ=\text{polar}(M)$. Since obviously $MQ=\text{polar}(P)$ by Brokard's Lemma $PM=\text{polar}(Q)$. Thus, $(Q, M; A, B)$ is harmonic so $P(Q, M; A, B)=P(Q, M; C, D)$ is harmonic. Since $PQ\parallel CD$ this implies $M$ is midpoint of $CD$, done.

http://www.artofproblemsolving.com/community/c6h60529p365585 Same problem ( IMO 1994 - P2)

Because $DBOM$ and $ACOM$ are cyclic, $O$ is the center of spiral similarly so $\triangle AOB \sim \triangle COD $ so $COD$ is isosceles and $M$ is the midpoint of $CD$.

can some one think of butterfly th.

One liner Let $X=AB\cap XP$ where $XP\|CD$ and now, as $PQ$ is polar of $M$, $QM$ is polar of $P\implies (\infty_{CD},M;C,D)\overset{P}{=}(Q,M;A,B)=-1$ completing the proof.

jayme wrote: Dear Mathlinkers, the perpendicular to OM through which point? Sincerely Jean-Louis Hi. perpendicular to OM through M...

Olemissmath wrote: From the point $P$ outside a circle $\omega$ with center $O$ draw the tangents $PA$ and $PB$ where $A$ and $B$ belong to $\omega$.In a random point $M$ in the chord $AB$ we draw the perpendicular to $OM$, which intersects $PA$ and $PB$ in $C$ and $D$. Prove that $M$ is the midpoint $CD$. Sorry for bumping such a old thread. But here in the problem we have $AMB$ as the Simpson's Line from $O$ to $\triangle{CDP}$ , which is also the Simpson's Line from $O$ to $\triangle{ABP}$ . Also we have if $CD$ intersects the $\omega$ at $C^{`},D^{`}$ . Then $CC^{`}$ and $DD^{`}$ are of same length . Try all these problems.

$\measuredangle OAC=\measuredangle OMC=90^{\circ}$ so $OMAC$ is cyclic, similarly $OMBD$ is cyclic. clearly $OA=OB$ and $\measuredangle OAC=\measuredangle OBD$. We have $\measuredangle OCA=\measuredangle OMA=\measuredangle OMB=\measuredangle ODB$ So $\triangle OCA\cong \triangle ODB$ and $CA=DB$. Now by Menelaus we have $\frac{CA}{AP}\frac{AB}{BD}\frac{DM}{MC}=1$ so $DM=MC$.