In order for this to be true, we don't need the information that $XY$ is a diameter. $NP$ is $N$-symmedian of $NCD$ so it's enough to prove that $AB$ is antiparallel to $CD$ . $\angle NBA=\alpha$ and $\angle NYX=\beta\Rightarrow \angle DNY=\alpha-\beta\Rightarrow \angle YCD=\alpha-\beta$. Since $N$ is midpoint of arc $XY$, $\angle NCY=\beta$ so $\angle NCD=\alpha$, as needed.

Hmm.... I used projectivity to kill the problem. Let $J$ be the intersection of $NP$ with the circle.Then obciously $CDJN$ is a harmonic quadrilateral.Thus $NC,ND,NJ$ and the tangent at $N$ to the circle form a harmonic pencil.In other words $NA,NB,NM$ and the tangent at $N$ to the circle form a pencil.But note that the tangent at $N$ is parallel to diameter $XY$ since $N$ is the midpoint of arc $XY$.So $M$ is the midpoint of $AB$,as desired.

Let $X'$ and $Y'$ be the intersections of the line $XY$ with the lines $PC$ and $PD$, respectively. Using Menelaus's theorem we get \[ \frac{X'C \cdot PN \cdot MA}{PC \cdot MN \cdot X'A} = 1 = \frac{Y'D \cdot PN \cdot MB}{PD \cdot MN \cdot Y'B} \implies \frac{X'C \cdot PD \cdot MA \cdot Y'B}{Y'D \cdot PC \cdot MB \cdot X'A} = 1 \] Since $PC$ and $PD$ are tanget to $\Gamma$, $PC=PD$. Also, $\angle X'AC = \angle NAO = 90^\circ - \angle CNO = 90^\circ - \angle NCO = \angle X'CA$, then $X'C=X'A$. Analogously $Y'D = Y'B$. Then the first result simply becomes $MA=MB$.

My solution : Let $ \omega_A $ be a circle passing through $ A, C $ and tangent to $ XY. $ Let $ \omega_B $ be a circle passing through $ B, D $ and tangent to $ XY. $ Since $ NA \cdot NC=NB \cdot ND, PC^2=PD^2, $ so $ NP $ is the radical axis of $ \omega_A $ and $ \omega_B $ $ \Longrightarrow $ $ MA^2=MB^2. $ Q.E.D

through P we drawn ST//AB (S on NC and T on NB) we have m(XAC)=m(XCA) (because NX=NY) ==> m(PCS)=m(PSC)==> PS=PC,analogously PD=PT ==> PS=PT and combinate ST//AB.

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Solution. By definition of $N$, we get $$NA\cdot NC=NX^2=NY^2=NB\cdot ND$$thus $ABDC$ is cyclic. Since $NP$ is the $N$-symmedian of $\bigtriangleup CND$, we're done. $\blacksquare$

Using some trig: We'll need the following lemma: Let $ABC$ a triangle and $AD$ a cevian. Then $\frac{\sin(\angle BAD)}{\sin(\angle DAC)} = \frac{BD}{DC}\frac{CA}{AB}$. Let $M'$ be the intersection between $CD$ and $PN$, and let $Q$ the intersection between the circumference and $CD$. We have that $\angle{ NAB} = \angle NDC$ since they both open the same arcs and $NX = NY$. Similarly, $\angle NBA = \angle NCD$ and then triangles $NAB$ and $NDC$ are similar and therefore $\frac{AN}{BN} = \frac{ND}{NC}$. Observe that $\angle CDQ = \angle CNQ = \angle PCQ$ since $\angle CDP = \angle PCD = \angle CND$, and similarly $\angle PND = \angle DCQ = \angle QDP$. Call $\alpha = \sin(\angle CNP), \beta = \sin(\angle PND)$ Trig version of Ceva on $CDP \implies \frac{\sin(\angle DPM')}{\sin(\angle M'PC} \frac{\alpha^2}{\beta^2} = 1 \implies \frac{\sin(\angle DPM')}{\sin(\angle M'PC} = \frac{\beta^2}{\alpha^2} = \frac{M'D}{M'C}$, where the last equality comes from using the lemma on triangle $CPD$ with cevian $PM'$ Now the lemma on triangle $CND$ and cevian $NM'$ gives $\frac{\alpha}{\beta} = \frac{CM'}{M'D}\frac{ND}{NC} \implies \frac{\beta}{\alpha} = \frac{ND}{NC} = \frac{AN}{BN}$. Finally, apply the lemma on $NAB$ with cevian $NM \implies \frac{\alpha}{\beta} = \frac{AM}{BM}\frac{BN}{AN} = \frac{AM}{BM}\frac{\alpha}{\beta} \implies \frac{AM}{BM} = 1$, as desired

Davi Medeiros wrote: Let $X$ and $Y$ be the diameter's extremes of a circunference $\Gamma$ and $N$ be the midpoint of one of the arcs $XY$ of $\Gamma$. Let $A$ and $B$ be two points on the segment $XY$. The lines $NA$ and $NB$ cuts $\Gamma$ again in $C$ and $D$, respectively. The tangents to $\Gamma$ at $C$ and at $D$ meets in $P$. Let $M$ the the intersection point between $XY$ and $NP$. Prove that $M$ is the midpoint of the segment $AB$. Posting for Storage. Have essentially the same solution as #8, but I spelled out the ending Let $K$ be the midpoint of $DC$ and let the tangents at $N$ and $C$ intersect at a point $T$ and let the tangents at $N$ and $D$ intersect at a point $S$. We know that the tangent through $N$ is parallel to $XY$ because $N$ is the midpoint of arc $XY$. By the tangent chord theorem, we know that $\angle BDC = \angle NDC = \angle CNT = \angle NAY = \angle NAB$ meaning that $\triangle NAB \sim \triangle NDC$. Moreover, by the Symmedian Lemma, $\angle KND = \angle PNC$. Then by the Ratio Lemma and using the fact that $\triangle NAB \sim \triangle NDC$ $$\frac{AM}{MB} = \frac{\sin \angle MAN}{\sin \angle MNB} \cdot \frac{AN}{BN} = \frac{\sin \angle DNK}{\sin \angle KNC} \cdot \frac{DN}{NC} = \frac{DK}{KC}$$where the last equality folows from the Converse of the Ratio Lemma. Clearly, by definition $\frac{DK}{KC} = 1$ meaning that $M$ is indeed the midpoint of $AB$.

Instat-killled by symmedians... In fact, we will prove the result in general for any chord $XY$. By definition, $NP$ is a symmedian in $\triangle NCD$. If we can show that $AB$ is anti-parallel to $CD$, then we know that the symmedian bisects all antiparallel lines, so we would be done. This is easy; just notice that $\angle XBN=\frac{\overarc{XN}+\overarc{YD}}{2}=\frac{\overarc{YN}+\overarc{YD}}{2}=\angle NCD,$ so we are done.

As $N$ is midpoint of arc $XY$ of $\Gamma$ we have$\angle NYX = \angle NXY = 45^{\circ}$. Then, $\angle NCD = \angle NCY + \angle YCD = 45^{\circ} + \angle YND = \angle NBX \Rightarrow ABDC$ is a cyclic quadrilateral. So $\triangle NAB \sim \triangle NDC$. Let $M'$ be the midpoint of $CD$. It's well know that $NP$ y $NM'$ are isogonals, then $\angle CNP = \angle ANM = \angle DNM'$, therefore $M$ is to the $\triangle NAB$ as $M'$ is to the $\triangle NDC$. Thus $M$ is midpoint of $AB$ $\blacksquare$

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