I note the intersection $R\in AM\cap DN$. We will analize the all radical axisis between the circumcircles of the (inscribed) quadrilaterals $AMPZ,\ DNPZ$ and the circumcircles with the diameters $AC,\ BD$ a.s.o.

I note: $C(XY)$ - the circle with the diameter $XY;$ $\ R\in AM\cap DN,\ R_1\in AM\cap XY,\ R_2\in DN\cap XY$; the circles $a=C(AC),\ b=C(BD),\ m=C(AP),\ n=C(DB)$; the radical axises $d_1,\ d_2$ for the pairs of the circles $(a,n),\ (b,m)$ respectively. The radical center of the circles $(a,b,m)$ is $R_1\in XY\cap AM\cap d_2\Longrightarrow R_1\in d_2$; The radical center of the circles $(a,b,n)$ is $R_2\in XY\cap DN\cap d_1\Longrightarrow R_2\in d_1$; The radical center of the circles $(a,m,n)$ is $R_1\in AM\cap XY\cap d_1\Longrightarrow R_1\in d_1$; The radical center of the circles $(b,m,n)$ is $R_2\in DN\cap XY\cap d_2\Longrightarrow R_2\in d_2.$ Therefore, $R_1\equiv R_2\in d_1\cap d_2\cap XY\cap AM\cap DN\Longrightarrow R_1\equiv R_2\equiv R\in XY.$

we know that ${XY}\perp{AD}$ at $Z$. let ${AM}\cap{XY} = R$. we have $\angle{RMC} = 90$ (AC is diameter) so, $RMCZ$ is cyclic, that implies $PR.PZ = PM.PC$. let ${ND}\cap{XY} = S$. analogous, we have $PS.PZ = PN.PB$. $XY$ is the radical axes of the two circles, so, $PM.PC = PN.PB \Longrightarrow PR = PS \Longrightarrow R = S$.

I view the four circles have the same eje radical

Suppose $E = AM \cap XY$ and $N' = ED \cap XY.$ We have $\angle AMC = 90^\circ.$ Let $\angle MAC = \alpha, \angle EDZ = \beta, \angle AEZ = 180^\circ - \alpha = \theta_1,$ and $\angle DEZ = 180^\circ - \beta = \theta_2.$ Now if we show that $\angle PBZ = \theta_2,$ we show that $N' = N$ and we are done. Since $Z$ lies on the radical axis $XY$ of the two given circles, $(AZ)(CZ) = (BZ)(ZD).$ It can easily be seen that $\triangle EAZ \sim \triangle CPZ$ from which we get $(EZ)/(CZ) = (AZ)/(ZP).$ Thus, $(EZ)(ZP) = (AZ)(CZ) = (BZ)(ZD),$ so $(EZ)/(BZ) = (ZD)/(ZP),$ so $\triangle BPZ \sim \triangle EDZ,$ and we are done.

Let $R = AM \cap DN$. $P$ is a point on radical axis of the circlec with diameter $AC$ and $BC$. So we have $PC\cdot PM = BP \cdot PN $ which implies $B,C,N,M$ are concyclic. That is $\angle NMC = \alpha = \angle NBD \Rightarrow \angle BDN = 90^{\circ}-\alpha$ and $\angle AMN=90^{\circ}+\alpha$ which also implies $A,M,N,D$ are concyclic. Thus $RM\cdot RA = RN \cdot RD$ which means $R$ is a point on the radical axis of those circles. So $R,X,P,Y$ are concurrent.

This message might beput away

Solution using Menelaus' theorem: Let $AM \cap XY = T$. Equivalently we have to prove that $T,N,D$ are collinear. By Menelaus' theorem $T,N,D$ will be collinear if and only if $\frac{ZT}{PT}\cdot \frac{PN}{BN} \cdot \frac{BD}{DZ} = -1$ Observe that $ \triangle AMC \sim \triangle PZC, \implies \frac{CA}{CM}= \frac{PC}{ZC} $ and $\triangle BND \sim \triangle PZB \implies \frac{PB}{ZB}= \frac{BD}{BN} $ . By intersecting chord theorem $ PB \cdot PN = PM\cdot PC $ and $ ZB \cdot ZD = ZA\cdot ZC $ . Note that $\frac{PM}{ZA}\cdot \frac{CA}{CM} = \frac{PM\cdot PC}{ZC \cdot ZA} = \frac{PM\cdot PC}{ZB\cdot ZA} = \frac{PB\cdot PN}{ZB \cdot ZD}= \frac{BD\cdot PN}{ZD \cdot BN }$. Since $T,M,A $ are collinear $\frac{ZT}{PT}\cdot \frac{PM}{CM} \cdot \frac{CA}{AZ} = \frac{ZT}{PT}\cdot \frac{PN}{BN} \cdot \frac{BD}{DZ} =-1$. $Q.E.D$

Let first establish that: (1) $\frac{ZA}{ZB}=\frac{ZD}{ZC}$. From the right triangle $AXC$ we get $ZA.ZC=ZX^2$. Similarly, from the right triangle $BXD$ it follows: $ZB.ZD = ZX^2$. Hence $ZA.ZC=ZB.ZD$, which implies (1). Now the homothety with center $Z$ and ratio $k=\frac{ZA}{ZB}$ will send the altitude through $B$ to $PC$, to the line $AM$; the altitude through $C$ to $BP$ will go to the line $DN$, and the line $ZP$(which is the altitude through $P$ to $BC$) will be sent to itself. This means that $AM, DN, ZP$ are concurent and its common point is the image of the orthocenter of $\Delta BCP$ by this homothety.

Let $AD \cap XY=L$ Notice that $ALOM$ and$NDLO$ are cyclic .So $\angle OMY=\angle MOY$ and $\angle YON=\angle YNO$. So O is the circumcentre of $\triangle OMN$.and thus by angle chasing we get $AMND$ cyclic.THus the lines$AM$ and$DN$ are meeting on the radical axis.

This problem is easy but tasty.... Notations:$AM \cap DN=K,\angle{BNM}=\theta,\odot{AMC}=\Omega_1,\odot{BND}=\Omega_2$. ProofSince $P$ lies on the radical axis $XY$ of $\Omega_1$ and $\Omega_2$,we have $BP*PN=CP*PM \Rightarrow B,C,N,M$ concyclic.So $\angle{MCB}=\angle{MNB}=\theta$.Hence $\angle{MAD}=90^{\circ}-\theta \Rightarrow M,A,D,N$ concyclic.Hence $KM*KA=KN*KD \Rightarrow K$ lies on the radical axis $XY$ as desired.

Let $Q=AM\cap XY$ and let $QD$ intersect the circle with diameter $BD$ at $D, N'$. We wish to show $N'\equiv N$. Note that \[\angle AMC =\angle QMP =\angle BND =\angle BN'D =\angle AZQ =\angle QN'P=\frac{\pi}{2}.\] Then $\angle QMP +\angle QN'P = \pi $, so $QMPN'$ is cyclic. Now $\angle MPN' =\angle BPC$ and \[\angle MN'P=\angle MQP =\frac{\pi}{2}-\angle QAC =\angle BCP,\] so $\triangle MN'P\sim \triangle BPC$. Hence, \[\frac{PM}{PB}=\frac{PN'}{PC}\implies PM\cdot PC=PB\cdot PN'.\] But since $P$ is on the radical axis, \[PM\cdot PC=PB\cdot PN\implies PB\cdot PN=PB\cdot PN'\implies PN=PN'.\] Hence $N\equiv N'$ and $AM,XY,DN$ concur as desired. $\blacksquare$

A very nice proof is also given here: https://www.math.ust.hk/excalibur/v5_n3.pdf

This is from the book! There's no radical axes or bashing.

The second paragraph of the solution is well-known.

We see $\angle AMC$ is $90$, and also $\angle PZA$ is $90$. Therefore, $AMPZ$ is cyclic. Similarly $NPZD$ is cyclic. Let $C_{1}, C_{2}, C_{3}, C_{4}$ denote the circumcircles of $AMC, DNB, AMPZ, DNPZ$ respectively. Now $AM$ is the radical axis of $C_{1}$ and $C_{3}$ and $DN$ is the radical axis of $C_{2}$ and $C_{4}$ and $XY$/$PZ$ is the radical axis of $C_{1},C_{2}$ and $C_{3},C_{4}$. Let $Q$ be the intersection of $AM$ and $XY$.Therefore at $Q$ power of all the circles are equal. Let $ND$ and $XY$ meet at a different point (say $H$).Then at $H$ too the power of all the circles is same.But $AM$ is the locus of all points which have the same power with circles $C_{1}$ and $C_{3}$.That implies $H$ lies on $AM$ and hence $Q$ and $H$ coincide.

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Denote $w_1$ and $w_2$ as the circles with diameters of $\overline{AC}$ and $\overline{BD}$ respectively. Since $P$ is on the radical axis of $w_1$ and $w_2$, quadrilateral $AMND$ is cyclic. Since $T=\overline{AM}\cap \overline{ND}$, and since $AMND$ is cyclic, we must have that $T$ is on $\overline{XY}$. The result follows.

Let the circle with diameter $AC$ be denoted as $\omega_1$ and the circle with diameter $BD$ be denoted as $\omega_2$. Lemma: Given two circles $\Omega_1$ and $\Omega_2$, line $AB$ and line $CD$, where $A$ and $B$ are points on $\Omega_1$ and $C$ and $D$ are points on $\Omega_2$ , concur on the radical axis of $\Omega_1$ and $\Omega_2$ if and only if $ABCD$ is cyclic. Proof: Let $AB$ and $CD$ intersect at point $P$. If $P$ lies on the radical axis, we have $Pow_{\Omega_1}(P)=Pow_{\Omega_2}(P)\implies PA\cdot PB=PC\cdot PD.$ If $ABCD$ is cyclic, we have $PA\cdot PB=PC\cdot PD$ by Power of the Point. Thus, this proves our if and only if statement.$\Box$ Claim 1: $MNCB$ is cyclic. Proof: This comes trivially from the Lemma applied to lines $MC$ and $BN$ of circles $\omega_1$ and $\omega_2$ respectively. $\Box$ Claim 2: $AMND$ is cyclic Proof: By Claim 1, note that $\angle NMC=\angle NBC$. Hence, $$\angle AMN+\angle NDA=90^{\circ}+\angle NMC+\angle NDA=90^{\circ}+\angle NBC+\angle NDA=180^{\circ},$$proving the desired. $\Box$ Since $AMND$ is cyclic, the result follows by the Lemma. $\blacksquare$

orl wrote: Let $ A,B,C,D$ be four distinct points on a line, in that order. The circles with diameters $ AC$ and $ BD$ intersect at $ X$ and $ Y$. The line $ XY$ meets $ BC$ at $ Z$. Let $ P$ be a point on the line $ XY$ other than $ Z$. The line $ CP$ intersects the circle with diameter $ AC$ at $ C$ and $ M$, and the line $ BP$ intersects the circle with diameter $ BD$ at $ B$ and $ N$. Prove that the lines $ AM,DN,XY$ are concurrent. This was a good problem. It suffices to prove that ANMD is cyclic, where the motivation comes from concurrency using radical center being that one axis is XY, the other two need to contain AM and DN so we look at ANMD. Now denote the intersection of AM with DN as E, and note the right angles ENP and EMP. We have <AMN=<EMN=90-<NMP=90-<NBC (pop stuff)=<ADN (since BND is a right triangle), so ADMN is cyclic, as desired.

Solution 1 (BMC). Let $E:=AM\cap XY$ and $F:=DM\cap XY$; we want to show $E=F$. Clearly $BCMN$ is cyclic because $PB\cdot PN=PX\cdot PY=PC\cdot PM$, and $BFNO,CMFO$ are cyclic by Thales'. Then $P$ is the radical center of $(BCMN),(BFNO),(CMFO)$ with $OP$ the radical axis of $(BFNO),(CMFO)$, meaning $(BFNO)\cap(CMFO)=\{E,F,O\}$ or $E=F$ as desired. $\square$ Solution 2 (EGMO). As before $BCMN$ is cyclic, and $ADMN$ is cyclic since \[\angle DAM+\angle DNM=\angle DAM+\angle BNM+90^\circ=\angle DAM+\angle BCM+90^\circ=180^\circ\]by Thales' twice. So if $Q=AM\cap DN$, we have $QA\cdot QM=QD\cdot QN$ and so $Q\in XY$, as desired. $\square$

Note that it suffices to prove that $AMND$ is cyclic. (By the Radical Center of Intersecting Circles theorem, or Theorem 2.9 from EGMO.) By Power of a Point in the circle with diameter $\overline{AC}$, we have $$XP \cdot PY = CP \cdot PM,$$and by Power of a Point in the circle with diameter $\overline{BD}$, we have $$XP \cdot PY = BP \cdot PN.$$ Hence, it follows that $$CP \cdot PM = BP \cdot PN,$$so the converse of Power of a Point tells us that quadrilateral $BMNC$ is cyclic. Now, we angle chase to get \begin{align*} \angle{MND} &= \angle{MNB} + \angle{BND} \\ &= \angle{MCB} + 90^\circ \\ &= (90^\circ-\angle{MAC}) + 90^\circ \\ &= 180^\circ - \angle{MAC}, \end{align*}which implies that $AMND$ is cyclic, as desired. $\square$

Let $T = AM \cap DN$. Since $XY$ is the radical axis of $(AC)$ and $(BD)$, it suffices to show that $T$ also lies on the radical axis of $(AC)$ and $(BD)$. Claim: $AMND$ is cyclic. Proof. Since $P \in XY$, $P$ lies on the radical axis of $(AC)$ and $(BD)$. It follows that $\text{Pow}_{(AC)} P = \text{Pow}_{(BD)} P$, or $$PM \cdot PC = PN \cdot PB$$Thus, $BMNC$ is cyclic. Further, since $AC$ is the diameter of $(AC)$ and $M$ lies on $(AC)$, we have $\angle AMC = 90^{\circ}$. By angle chasing, we have \begin{align*} \angle MAD &= 90^{\circ} - \angle MCA \\ &= 90^{\circ} - \angle MNB \\ \end{align*}However, we also have $\angle MND = 90^{\circ} + \angle MNB$, which leads to $\angle MND + \angle MAD = 180^{\circ}$, and the conclusion follows. $\blacksquare$ Then, by Power of a Point, we have $TM \cdot TA = TN \cdot TD$, which is equivalent to $T$ lying on the radical axis of $(AC)$ and $(BD)$, as desired. $\blacksquare$

Note that $\operatorname{Pow}_{(AC)} (P) = \operatorname{Pow}_{(BD)} (P)$, so $BNMC$ is cyclic. Hence, \[\angle MAD = 90^\circ - \angle MCB = \angle MNB - 90^\circ = \angle MND,\] which implies $ANMD$ is cyclic. Then, note that $Z$ lies on $(AP)$ and $(DP)$, and using the radical center of $(AP)$, $(DP)$, and $(ANMD)$ concludes.

Let the point $Q$ be the intersection of $AM$ and $DN$. We use directed angles $\measuredangle$ modulo $\pi$. It suffices to then show that $MNDA$ is cyclic, as then $QM \cdot QA = QN \cdot QD,$ which would imply $Q$ has the same respective power to both circles and consequently $Q, X, Y$ are collinear. Make the first observation that $QMPN$ is cyclic as \[\measuredangle QMP = \measuredangle QMC = 90^{\circ} = \measuredangle QNB = \measuredangle QNP.\]We subsequently discover \begin{align*} \measuredangle QMN &= 90^{\circ} - \measuredangle NMC \\ &= 90^{\circ} - \measuredangle NBC \\ &= 90^{\circ} - \measuredangle NBD \\ &= \measuredangle BDN, \end{align*}so that indeed $MNDA$ is cyclic, as was to be shown. $\blacksquare$

Can cord-bash work??

Claim: Denote $AM \cap DN =K$ then, $K$ lies on the line $XY$. Proof: Denote the circumcircles of $\Delta AMC$ and $\Delta BND$ as $\Omega_{1}$ and $\Omega_{2}$. Note that $XY$ is the radical axis of $\Omega_{1}$ and $\Omega_{2}$. Thus by Power of Point, $BP.PN=CP.PM$ which implies that points $B,M,N,C$ are concyclic. As $AC$ is the diameter of $\Omega_{1}$ and $BD$ is the diameter of $\Omega_{2}$, we have $\angle AMC=\angle BND=90^{\circ}$. Thus, $90^{\circ}-\angle MAC=\angle MCB=\angle MNB \implies \angle MND= \angle MNB+\angle BND=180^{\circ}-\angle MAD$. Hence, points $A,M,N,D$ are concylic and as $K=AM \cap DN$ which implies, $KM.KA=KN.KD$ or the power of $K$ with respect to $\Omega_{1}$ and $\Omega_{2}$ is same $\implies K \in XY$ as desired. $\blacksquare$

Since $AC$ and $BD$ are diameters, $\measuredangle AMC=90$ and $\measuredangle BND=90$ Claim:$MCBN$ is cyclic Proof: $XP\times PY=BP\times PN$ $XP\times PY=MP\times PC=BP\times PN\Rightarrow MCBN$ cyclic $\measuredangle MND=\measuredangle MNB+\measuredangle BND=90+\measuredangle MNB=90+\measuredangle MCB=90+\measuredangle MCA$ $\measuredangle MCA+\measuredangle AMC+\measuredangle CAM=180\Rightarrow \measuredangle CAM=90-\measuredangle MCA\Rightarrow \measuredangle MAC=\measuredangle MCA-90\Rightarrow \measuredangle MAD=\measuredangle MCA-90=\measuredangle MCA+90$ $\measuredangle MAD=\measuredangle MCA+90=\measuredangle MND\Rightarrow MAND$ cyclic. Circle $MAND$ and $AMCY$ intersects at $A;M\Rightarrow AM$ is the radical axis of these 2 circles. Circle $MAND$ and $BNDY$ intersects at $B;N\Rightarrow BN$ is the radical axis of these 2 circles. Circle $AMCY$ and $BNDY$ intersects at $X;Y\Rightarrow XY$ is the radical axis of these 2 circles. SInce $P\neq Z$,center of these 3 circles aren't colinear,so their radical axes($AM;BN;XY$) concur.

$XP\cdot PY=BP\cdot PN=CP\cdot PM$ gives $BCMN$ concyclic. Recall that $AC$ and $BD$ was diameters. Hence, $\angle MAC=90^{\circ}-\angle MCA=90^{\circ}-\angle MNB$ implies $ADMN$ concyclic. Then, $AM$, $DN$ and $XY$ are concurrent as desired.

Let the circle with diameter $AC$ be $\omega_1$ and the circle with diameter $BD$ be $\omega_2$. Let $Q$ be the intersection of $AM$ and $XY$ and $N'$ be the intersection of $QD$ with $\omega_2$. Since $AC$ is the diameter of $\omega_1$, $\angle AMC = \angle QMC = 90$. Since $XY$ is the radical axis of $\omega_1$ and $\omega_2$, $\angle QZD = \angle QZA = 90$ so $\triangle QMP \sim \triangle QZA$ by AAA. Hence $\frac {QM}{QZ} = \frac {QP}{QA}$ so $QM\times QA = QP\times QZ$. From power of a point, $QM\times QA = QN'\times QD$ so $QP\times QZ = QN'\times QD$. Hence $\frac{QP}{QD} = \frac{QN'}{QZ}$ so $\triangle QN'P \sim \triangle QZD$, so $\angle QN'P = 90 = \angle PN'D$ as $QN'D$ is a straight line. Thus $PZDN'$ is cyclic so the radical axis of $PZDN'$ and $\omega_2$ is $DN'$. However $PZDN$ is also cyclic (since $\angle BND = \angle PND = 90$) so the radical axis of $PZDN$ and $\omega _2$ is $DN$. Since two circles cannot have more than one radical axis ($PZDN$ and $PZDN'$ are the same circles), $N'$ must lie on $DN$. But since $N$ and $N'$ lie on $\omega_2$, $N=N'$ so $AM$, $DN$ and $XY$ are concurrent