Let $ X,Y,Z$ be the midpoints of the small arcs $ BC,CA,AB$ respectively (arcs of the circumcircle of $ ABC$). $ M$ is an arbitrary point on $ BC$, and the parallels through $ M$ to the internal bisectors of $ \angle B,\angle C$ cut the external bisectors of $ \angle C,\angle B$ in $ N,P$ respectively. Show that $ XM,YN,ZP$ concur.
Problem
Source: IMO Shortlist 1997, Q25
Tags: geometry, circumcircle, incenter, parallelogram, concurrency, IMO Shortlist
25.09.2004 17:15
I know the source. It's the IMO Shortlist 1997. It's on Kalva's ISL page, but you needn't click on that link, it doesn't contain the solution. I can recall me and Peter (Scholze) killing this problem algebraically on the IMO (more precisely, during the Opening Ceremony of the IMO ), but I don't have a synthetic solution. Darij
25.09.2004 18:56
yeah, we spent some time trying to use every theorem giving concurrence of three lines as a result, but we didn't see anything(anyway, it was quite useful for me since i learnt many useful theorems which i could have used for IMO, but there was (unfortunately) no problem to use them). our algebraic proof, anyway, is still very nice. we just proved it for 4 points M, which suffices since the condition is a degree 3-equation Image not found Peter
26.09.2004 23:41
If we turn this problem around : make a circumcircle and make a point P move along an external angle bisector of B, then a line from P thru a fixed point on the circumcircle (mid-arc point Z) will have second intersection W, and the line from N parallel to the 'other' internal bisector C intersects a fixed line BC at M then line WM goes thru a fixed point X(mid-arc of BC). Selecting mid-arc fixed points insures syncronized move of point N along the external angle bisector of C. A circle thru points B,P,W,M proves that <PWM = <PBA . Pestich. I wonder if it is possible to construct a triangle that has the circumcircle of ABC as it's NPC, so that two Simson lines of antipod points intersect on it?
27.09.2004 14:45
Pestich, you probably remember APH answering to one of your messages on Hyacinthos. He quoted your text and commented: "Probably only Darij could decipher the above." He was wrong, I could not. In most other cases, I can, but it takes me an hour to understand what you intend to say. Could you please express yourself in a more detailed way? The way you write your ideas is very poetic, but it is very hard to understand the mathematical contents. Concerning the problem: Now I do have a synthetic solution. It is quite nice, it doesn't use any of the hardcore theorems I told Peter in Athens, and it proves even more, namely that the point of concurrence of the lines XM, YN and ZP lies on the circumcircle of triangle ABC. But the few minutes I have left before another boring school lesson starts are definitely not enough to post it. I will come back this evening (I hope). Let's say, with Hxtung: "I'll post my soln". Darij
27.09.2004 18:52
It seems that this problem is a bit interesting. Ok I 'll try it after school. PS Darij, you know Russian ? I wonder if that journal has the English version. It seems not.
27.09.2004 18:58
If I is the incenter of ABC then xAI, YBI and ZCI. Is not difficult to prove that XY is ortogonal to CI then XY//CN and similarly XZ//PB. If U is the second common point to c(ABC) and c(APN) we have that P,Z,U are collinear. Infact <UZX = 180-<UAX. Since PB and Ax meet in Q on c(APN) [not difficult to prove] it holds that <UPQ = 180-<UAQ=180-<UAX. Then <UZX = <UPQ from which, as PB//ZX, we have the claim. Similarly one can proove that N,Y,U are collinear. I will prove now that X,M,U are collinear too. It holds that <YUB = <YAB =<YCB and, as CY//MN, <YCB = <NMB then NBMU is ciclic. But NM meet AC in O on c(ANP) and then <BNM = <QNO = <QAO = <XAC = <XAB = <XUB which implies the thesis.
27.09.2004 20:51
Sorry for 'dyslexic' mix-up of points P and N. It was not intentional and it is fixed now. Darij, please do not be so touchy about my a bit controversial remark some time ago. A little controversy is good. In the problem I just wanted to point out the similarities of this one and Treegoner's in this section ' Lying on a fixed line' . Here a ray from arbitrary point P on the external angle bisector of B thru fixed point Z on a circle is bent into another fixed point on a circle (X). How about a triangle that has 2 Simson lines intersecting on it's 9-PC ? Can we make lines PZ and NY to be Simson lines of some triangle? Pestich.
27.09.2004 22:07
Pestich, it wasn't at all about the controversy, it was about the a bit hard to understand mathematical style. Indeed, I cannot judge your ideas about the "Lying on the fixed line" configuration, since I admit I didn't really study it (just too many points on a single sketch!). What concerns the Simson lines, actually the circumcircle of triangle ABC is the nine-point circle of the excentral triangle, but the lines PZ and NY are (unfortunately) not Simson lines with respect to the excentral triangle. However, the idea is nice: indeed it looks like there are some Simson lines lurking around... Treegoner, which journal?! (Elemente der Mathematik?) Grobber, one little note: You wrote: "Let X, Y, Z be the midpoints of the small arcs BC, CA, AB respectively (arcs of the circumcircle of ABC)." This definition doesn't work for obtuse-angled triangles ABC; a definition working for every triangle would be: "Let X, Y, Z be the midpoints of the arcs BC, CA, AB not containing A, B, C, respectively (arcs of the circumcircle of ABC)." And here is the promised solution (even if Sprmnt21 already posted another one, which I, however, don't understand - probably a few typos). Peter, if you are interested: Just read the formulation of Lemma 1, that of Lemma 2, maybe the Note and a part of the proof of Lemma 2, and most importantly the last paragraph; all the rest is just completely trivial (but there are many people asking for such things, and now, at least, I will be able to point them to this posting). I will work with directed angles modulo 180. At first, I recapitulate a lemma: Lemma 1. Let $I_a$, $I_b$, $I_c$ be the excenters of triangle ABC. Then, the triangles $I_aBC$, $AI_bC$ and $ABI_c$ are all directly similar to each other, and their circumcenters are the points X, Y, Z. The points X, Y and Z also coincide with the midpoints of the segments $II_a$, $II_b$, $II_c$. Note. Please note the order of the vertices when I say "the triangles $I_aBC$, $AI_bC$ and $ABI_c$ are all directly similar to each other". When I say that two triangles PQR and P'Q'R' are similar, then I mean that P and P' are corresponding vertices, that Q and Q' are corresponding vertices, and that R and R' are corresponding vertices. Proof. Let I be the incenter of triangle ABC. The lines AI, BI, CI are the angle bisectors of the angles CAB, ABC and BCA, respectively, and hence they pass through the points X, Y, Z (since, as the point X is the midpoint of the arc BC of the circumcircle of triangle ABC, the chords BX and CX are equal, and thus the chordal angles of these two chords are equal, too, i. e. we have < BAX = < XAC, so that the point X lies on the angle bisector of the angle CAB, and similarly the points Y and Z lie on the angle bisectors of the angles ABC and BCA). These lines AI, BI, CI also pass through the excenters $I_a$, $I_b$, $I_c$. The lines $I_bI_c$, $I_cI_a$, $I_aI_b$ pass through the points A, B, C, respectively, and are the external angle bisectors of the angles CAB, ABC, BCA, respectively. Since an internal angle bisector of an angle is perpendicular to the external angle bisector, we have $\measuredangle IBI_a=90^{\circ }$ and $\measuredangle ICI_a=90^{\circ }$. Thus, the points B and C lie on the circle with diameter $II_a$. Consequently, $\measuredangle I_{a}BC=\measuredangle I_{a}IC$ and $\measuredangle BCI_{a}=\measuredangle BII_{a}$. Similarly, the points C and A lie on the circle with diameter $II_b$; thus, $\measuredangle AIC=\measuredangle AI_{b}C$ and $\measuredangle I_{b}IA=\measuredangle I_{b}CA$. Hence, $\measuredangle I_{a}BC=\measuredangle I_{a}IC=\measuredangle AIC=\measuredangle AI_{b}C$; $\measuredangle BCI_{a}=\measuredangle BII_{a}=\measuredangle I_{b}IA=\measuredangle I_{b}CA$. Hence, the triangles $I_aBC$ and $AI_bC$ are directly similar. Similarly, the triangles $AI_bC$ and $ABI_c$ are directly similar. Thus we have proven the direct similarity of triangles $I_aBC$, $AI_bC$ and $ABI_c$. [Of course the above was trivial, but I wanted to show how a real angle chase looks like ;-) ] Now, since the points B and C lie on the circle with diameter $II_a$, the circumcenter X' of triangle $I_aBC$ is the midpoint of this circle with diameter $II_a$, i. e. the midpoint of the segment $II_a$. Hence, this circumcenter X' must lie on the line AI. On the other hand, the point X', as the circumcenter of triangle $I_aBC$, must lie on the perpendicular bisector of the side BC. Therefore, the point X' is the point of intersection of the line AI with the perpendicular bisector of the side BC. But on the other hand, we know that the point X lies on the line AI, and moreover, the point X lies on the perpendicular bisector of the side BC, since X is the midpoint of the arc BC. Hence, the point X can also be characterized as the point of intersection of the line AI with the perpendicular bisector of the side BC. Consequently, X' = X, what implies that the point X is the circumcenter of triangle $I_aBC$ and the midpoint of the segment $II_a$. Similarly, the point Y is the circumcenter of triangle $AI_bC$ and the midpoint of the segment $II_b$, and the point Z is the circumcenter of triangle $ABI_c$ and the midpoint of the segment $II_c$. This completes the proof of Lemma 1. Now, time for another lemma: Lemma 2. Let x, y, z be three corresponding lines in the similar triangles $I_aBC$, $AI_bC$ and $ABI_c$, respectively, and let these lines x, y, z pass through the respective circumcenters X, Y, Z of these triangles. Then, the lines x, y, z concur at one point on the circumcircle of triangle ABC. Note. Two lines are called "corresponding" in two similar triangles if the similitude transformation mapping the first triangle to the second one maps the first line to the second line. In other words, the first line plays the same role in the first triangle as the second line in the second triangle. For instance, if we have two similar triangles and their respective Euler lines, then these Euler lines are two corresponding lines in these two similar triangles. The most prominent particular case of Lemma 2 is the fact that the Euler lines of triangles $I_aBC$, $AI_bC$ and $ABI_c$ concur at one point on the circumcircle of triangle ABC. Proof. Let the line x meet the circumcircle of triangle ABC at a point F (apart from the point X). It will be enough to show that this point F also lies on the lines y and z. Corresponding lines in directly similar triangles form equal angles. Thus, since the lines x and y are corresponding lines in the directly similar triangles $I_aBC$ and $AI_bC$, respectively, we have $\measuredangle \left( x;\;I_{a}B\right) =\measuredangle \left( y;\;AI_{b}\right) $. Hence, $\measuredangle \left( x;\;y\right) =\measuredangle \left( x;\;I_{a}B\right) +\measuredangle \left( I_{a}B;\;AI_{b}\right) +\measuredangle \left( AI_{b};\;y\right) $ $=\measuredangle \left( x;\;I_{a}B\right) +\measuredangle \left( I_{a}B;\;AI_{b}\right) -\measuredangle \left( y;\;AI_{b}\right) $ $=\measuredangle \left( I_{a}B;\;AI_{b}\right) =\measuredangle I_{a}I_{c}I_{b}$. On the other hand, < XFY = < XZY (since the point F lies on the circumcircle of triangle ABC). Since the points X, Y, Z are the midpoints of the segments $II_a$, $II_b$, $II_c$, we have $XZ\parallel I_{a}I_{c}$ and $YZ\parallel I_{b}I_{c}$, so that $\measuredangle XZY=\measuredangle \left( XZ;\;YZ\right) =\measuredangle \left( I_{a}I_{c};\;I_{b}I_{c}\right) =\measuredangle I_{a}I_{c}I_{b}$. Altogether, we conclude $\measuredangle XFY=\measuredangle XZY=\measuredangle I_{a}I_{c}I_{b}=\measuredangle \left( x;\;y\right) $. But < XFY = < (x; YF), and therefore we have < (x; YF) = < (x; y). This means that the lines YF and y are parallel, and since they have a common point (namely, the point Y), they must coincide, and the point F lies on the line y. Similarly, the point F also lies on the line z. And Lemma 2 is proven. Now, Lemma 2 makes clear that if we succeed to show that the lines XM, YN and ZP are corresponding lines in the similar triangles $I_aBC$, $AI_bC$ and $ABI_c$, then it follows that these lines XM, YN and ZP concur and that their point of concurrence lies on the circumcircle of triangle ABC. This is even more than the problem required to show. In the following, we will prove that the lines XM, YN and ZP are corresponding lines in the similar triangles $I_aBC$, $AI_bC$ and $ABI_c$. The lines $BI_b$ and MN are both perpendicular to the line $I_cI_a$; hence, $BI_b \parallel MN$, and we have by Thales $\frac{BM}{MC}=\frac{I_bN}{NC}$. In other words, the points M and N divide the corresponding sides BC and $I_bC$ of the similar triangles $I_aBC$ and $AI_bC$ in the same ratio. Consequently, the points M and N are corresponding points in these two triangles. On the other hand, the points X and Y are corresponding points in these two triangles (being the circumcenters of these triangles). Thus, the lines XM and YN are corresponding lines in these two triangles; similarly, the lines XM and ZP are corresponding lines in the triangles $I_aBC$ and $ABI_c$. And thus, altogether, the lines XM, YN and ZP are corresponding lines in the triangles $I_aBC$, $AI_bC$ and $ABI_c$. And we are done. Darij
28.09.2004 12:11
darij grinberg wrote: And here is the promised solution (even if Sprmnt21 already posted another one, which I, however, don't understand - probably a few typos). Darij Ciao Darij, you are very polite with me supposing that are only typos. Actually I write down a very bad sketch. I wiil attach a picture which, I hope, make clearer the various statement. Together with this figure, I will try to give reason of all (not trivial) sentences I wrote down.
28.09.2004 13:17
The style doesn't matter, actually I don't need the solution to be written in a very detailed way. But when the style is brief and moreover, there are typos, then it gets difficult to understand. For instance, you write: "Since PB and Ax meet in Q on c(APN) [not difficult to prove]". The lines PB and AX (I assume that you mean X instead of x) meet in $I_a$ - and $I_a$ doesn't lie on c(APN). There must be a typo. But where? Thanking in advance for the figure... Darij
28.09.2004 14:15
Darij! I hope your solution is correct. It is too boring to read solution which takes two screen pages.
28.09.2004 14:42
darij grinberg wrote: For instance, you write: "Since PB and Ax meet in Q on c(APN) [not difficult to prove]". The lines PB and AX (I assume that you mean X instead of x) meet in $I_a$ - and $I_a$ doesn't lie on c(APN). There must be a typo. But where? Thanking in advance for the figure... Darij Here is the figure. But I have to review me reasoning to make all clear. I perhaps found a way which make use of Pascal theorem.
Attachments:

29.09.2004 01:50
Your solution is marvelous, Darij. I knew your lemma but I didn't recognize it. It is a little bit difficult to notice. Another solution is indicated in my book. It also proves that they are concurent at a point lying on the circumcircle of ABC. First, call K the intersection of XM and (ABC). Denote N the intersection of $YK$ and $I_aI_b$. We 'll prove that NM is parallel to BY by proving KNCM is cyclic (which is not difficult). But you can also approach this problem by using the $xy$ coordinate though it is a buffalic (it made me sick )) : If you call T a point such that MPTN is a parallelogram, then T lies on $I_bI_c$. Then you choose the coordinate system $Oxy$ such that y'Oy is the line $I_aA$, x'Ox is the line $I_bI_c$. Here is my conjecture : If you call X, Y, Z are points lying on $I_aH, I_bH, I_c H $ such that XYZ and ABC are homothetic (which means $ XY//I_bI_a, YZ // I_bI_c, ZX//I_cI_a$), then show that XM, YN, PZ are concurent. It seems right unless my diagrams are wrong.
29.09.2004 10:19
I will try to rewrite in a cleare way what I say before. I am referring to the figure attacched to the previous message. Firts of all a simple fact: X,Y,Z stay on AI,BI and CI respectively, where I is the incenter of ABC. If Q=PB^NC it holds that AX pass through Q. This can be prooved in different ways. One is to observe that BICQ is ciclic and X and IQ are ithe center and a diameter of the circumcircle. Now if define R=PM^AC and T=NM^AB, it holds that ATQR is ciclic. Indeed it results that BQ and CQ are the perpendicular bisectors of TM and RM respectively. From this is easy to see that <TQM +<MQR =<B+<C and then the claim is proved. As PR//ZC then <PRA = <ZCA = <ZCB = <IQB = <PQA then Q lies on c(ATQR) for the same reasons N lies on c(ATQR). If U is the second common point to c(APN) and c(ABC), it holds that <AUY = <AXY = <AQN = <AUN which means that U,Y and N are collinear. Similarly one can prove that U,Z and P are collinear. It holds that <UNM = <UYB = <UAB = <UCB = <UCM which means that ANCM is ciclic. From which we have that <MUC = <MNC = <BYX = <BUX = <XUC in other words U,M and X are collinear too. I hope this time is all clear and perhaps correct.
29.09.2004 21:52
Myth wrote: Darij! I hope your solution is correct. It is too boring to read solution which takes two screen pages. Sorry, that's an old bad habit of mine. I always write down every trivial detail of the proof, resulting in five pages long proofs which nobody dares to read. I'll try to exercise a bit more in writing compactly. But anyway, I have forgotten to say that what I wrote to Peter applies to everybody halfways experienced in plane geometry: All what you need to read is Lemma 1 (without proof), Lemma 2 (with the Note), maybe a part of the proof of Lemma 2, and the very last paragraph of the solution (this very last paragraph contains the crux of the solution). Nothing more. That's my idiotic style. treegoner wrote: If you call X, Y, Z are points lying on $I_aH,I_bH,I_cH$ such that XYZ and ABC are homothetic (which means $XY//I_bI_a, YZ//I_bI_c, ZX//I_cI_a$, then show that XM, YN, PZ are concurent. I think you have confused something here, haven't you? I guess you considered the configuration "from the viewpoint of triangle $I_aI_bI_c$" and then slipped back to triangle ABC. But well, you seem to ask whether the lines XM, YN, ZP still concur if XYZ is an arbitrary triangle homothetic to triangle $I_aI_bI_c$ with homothetic center I. Yes, they are. How to prove this? Well, let's use, like Peter Scholze and Grobber, the Steiner theorem (I don't seem to have a better proof up to now). We parametrize the points X, Y, Z through the projective-real parameter $t=\frac{IX}{II_a}=\frac{IY}{II_b}=\frac{IZ}{II_c}$. ("Projective-real" is my notation for $\in \mathbb{R}\cup \left\{ \infty \right\} $.) Then, the concurrence of the lines XM, YN, ZP is proven for $t=\frac12$ (this was the problem), trivial for t = 0 (although I am not sure whether one is allowed to use such degenerate cases in Steiner's theorem), trivial for t = 1, almost trivial for $t=\infty$, and thus, if I am not wrong, these four values should imply the concurrence of the lines XM, YN, ZP for every t. Well, I am not really sure whether three values would be enough, or whether degenerate values can be used at all (degenerate cases are not necessarily limiting cases!); well, let's wait until somebody clarifies this. (BTW, is there a proof of the Steiner theorem without algebra and without conics?) A more down to earth proof of this all would be nice, of course. Sprmnt21, thanks for the solution! sprmnt21 wrote: Firts of all a simple fact: X,Y,Z stay on AI,BI and CI respectively, where I is the incenter of ABC. Yes, that was my Lemma 1 ;-) sprmnt21 wrote: If Q=PB^NC it holds that AX pass through Q. This can be prooved in different ways. Indeed, the simplest way is to realize that Q is the A-excenter of triangle ABC. sprmnt21 wrote: Now if define R=PM^AC and T=NM^AB, it holds that ATQR is ciclic. Indeed it results that BQ and CQ are the perpendicular bisectors of TM and RM respectively. From this is easy to see that <TQM +<MQR =<B+<C and then the claim is proved. As PR//ZC then <PRA = <ZCA = <ZCB = <IQB = <PQA then Q lies on c(ATQR) for the same reasons N lies on c(ATQR). If U is the second common point to c(APN) and c(ABC), it holds that <AUY = <AXY = <AQN = <AUN which means that U,Y and N are collinear. Similarly one can prove that U,Z and P are collinear. It holds that <UNM = <UYB = <UAB = <UCB = <UCM which means that ANCM is ciclic. From which we have that <MUC = <MNC = <BYX = <BUX = <XUC in other words U,M and X are collinear too. A very interesting proof. Sorry for my claim about a typo in your sentence "Since PB and Ax meet in Q on c(APN) [not difficult to prove]"; there is actually no typo here (I just was somehow irritated that you denoted $I_a$ as Q and drew a wrong sketch, from which I mistakenly concluded that there was a typo). Darij
30.09.2004 01:23
I also thought like you Darij. But is there any formal theory which states that if you have 4 values of t satisfying the conditions ? Sorry for my ignorance, what Steiner 's theory did you indicate ? About the journal Kbant, I just ask if you know Russian well because I don't think there is an English version of it.
30.09.2004 09:41
treegoner wrote: I also thought like you Darij. But is there any formal theory which states that if you have 4 values of t satisfying the conditions ? Sorry for my ignorance, what Steiner 's theory did you indicate ? I don't really know. Probably it would be better to use the identity theorem for polynomials like Peter did, but I am not sure whether it is allowed to use degenerate cases in such considerations. Anyway it's not a question to me. treegoner wrote: About the journal Kbant, I just ask if you know Russian well because I don't think there is an English version of it. Hmm, yes, I know Russian and I am from Russia. Some people claim there is an English version, but I have never seen it... Darij
30.09.2004 10:37
darij grinberg wrote: ... you denoted $I_a$ as Q and ... Darij Darij, I didn't see that Q was the a-excenter untill read yours messages . But I have a generalization of the problem where (at first insight) my reasonment seems yet working. Here is the claim: Let ABC be a triangle and P a point of the plain where ABC lies. Let X,Y and Z be the projections of A,B and C from P on c(ABC) and Q be a point s.t. QBC is similar to XYZ. If L is a point on BC let M be the intersection between the parallel through L to BP and CQ and let N be the intersection between the parallel through L to CP and BQ, the XL, YM and ZN are concurrent.
30.10.2010 22:55
09.11.2010 14:02
I think I found a short solution. Let $BP$ intersect $CZ$ at $E$, the excenter of $\triangle ABC$ opposite $C$. Let $XM$ intersect $PZ$ at $D$. Let $I$ be the incenter, then $Z$ is the midpoint of $EI$ and $\angle ZBE=\angle ZEB=\angle IAB=\angle BCX=\angle CBX$, so $\triangle ZEB\sim\triangle XCB$ and furthermore, $\frac{BM}{MC}=\frac{BP}{PE}$. Thus $\angle PZB=\angle BXM$ so $ZBXD$ is cyclic. Thus $D$ is the intersection of $XM$ with the circumcircle. Since $PZ$ passes through $D$ and similarly $NY$ passes through $D$, we have $PZ,XM,NY$ are concurrent (at $D$).
09.11.2010 18:30
Dear Mathlinkers. I cannot resist to use the Reim's theorem... 1. (0) the circumcircle of ABC, D, D' the second points of intersection of (0) with PZ, NY resp. 2. Remark : MP //CZ. 3. According to a converse of Reim's theorem, D, B, M, P are concyclic on a circle noted (1). 4. Mutatis mutandis, we would prove that D', C, M, N are concyclic on a circle noted (2). 5. Remark : BP//XZ 6. According to a converse of Reim's theorem applied to (0) and (1), BX is tangent to (1) at B. 7. Mutatis mutandis, we would prove that CX is tangent to (2) at C. 8. According to the pivot theorem applied to the triangle BCX, (0), (1) and (2) are concurrent ; consequently, D=D'. 9. Mutatis mutandis, we would prove that MX goes through D and we are done. Sincerely Jean-Louis
21.09.2012 00:00
Let $W$ be intersection of the circumcircle $(ABC)$ and $XM$. Then $\angle MPB= 180-\angle PMB -\angle PBM = 180-C/2- (B/2+90)=A/2=\angle BWX$, so $PBMW$ is cyclic. Then $\angle PWB =\angle PMB = C/2 =\angle ZWB$, so $P,Z,W$ are collinear. Similarly for $Y,W,N$.
Attachments:
1997 N20.pdf (154kb)
14.05.2014 12:05
I am looking for a solution using radical axis.I observed that $\angle{XBC}=\angle{BPM}=\frac{A}{2} \Rightarrow XB$ is tangent to $\odot{PBM}$.In a similar way $XC$ is tangent to $\odot{MNC}$.But $XB=XC \Rightarrow XN$ is the radical axis of the two circles.I only need another circle to prove the fact.Can anyone find out the third circle?? polya78 your solution is nice.
24.09.2021 00:19
Storage. Let $D=XM\cap (ABC)$. Let $N'$ be the intersection of line through $M$ parallel to $BI$ and $DY$, where $I$ is the incenter of $\triangle ABC$, we claim that $Y\equiv Y'$, i.e. $Y'$ lies on the $C$ external angle bisector. Firstly, $DMCN'$ is cyclic as $\measuredangle CDN'=\measuredangle CBY=\measuredangle CMN'$. Thus, $$\measuredangle MDN'=\measuredangle XDY=\measuredangle CBI+\measuredangle IAC=90^{\circ}-\measuredangle ICB=\measuredangle MCN'.$$Similarly, we get that $PZ$ passes through $D$, we are done.