Circle center O passes through the vertices A and C

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

orl

3647 posts

orl

#1 h Nov 11, 2005, 8:48 PM • 6 Y

Y by Davi-8191, Adventure10, Mango247, ehuseyinyigit, Rounak_iitr, cubres

A circle with center $O$ passes through the vertices $A$ and $C$ of the triangle $ABC$ and intersects the segments $AB$ and $BC$ again at distinct points $K$ and $N$ respectively. Let $M$ be the point of intersection of the circumcircles of triangles $ABC$ and $KBN$ (apart from $B$ ). Prove that $\angle OMB=90^{\circ}$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

darij grinberg

6555 posts

darij grinberg

#2 h Nov 13, 2005, 11:12 AM • 5 Y

Y by Amir Hossein, Adventure10, Mango247, cubres, and 1 other user

This problem was discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=84104 , http://www.mathlinks.ro/Forum/viewtopic.php?t=47101 and http://www.mathlinks.ro/Forum/viewtopic.php?t=549 (where the problem is given in a slightly modified version). However, let me give a different solution:

Let U and V be the circumcenters of triangles ABC and KBN, respectively.

We will use directed angles modulo 180°. Since the point V is the circumcenter of triangle KBN, i. e. the center of a circle passing through the points K, B and N, the central angle theorem for directed angles modulo 180° yields < KBV = 90° - < BNK. On the other hand, since the points C, A, K and N lie on one circle, < CNK = < CAK. Thus,

< (AB; BV) = < KBV = 90° - < BNK = 90° - < CNK = 90° - < CAK = 90° - < (CA; AB),

so that < (AB; BV) + < (CA; AB) = 90°, what simplifies to < (CA; BV) = 90°. Thus, $BV\perp CA$ . On the other hand, since the center of a circle lies on the perpendicular bisector of each chord of the circle, the points U and O both lie on the perpendicular bisector of the segment CA (in fact, the point U is the center of the circumcircle of triangle ABC, and the segment CA is a chord in this circumcircle; the point O is the center of the circle through the points C, A, K and N, which also has CA as chord; so both points U and O lie on the perpendicular bisector of the segment CA). Hence, the line UO is the perpendicular bisector of the segment CA (the line UO really exists, i. e. the points U and O don't coincide - else, we would have K = N = B, contradicting the problem condition). Thus, $UO\perp CA$ . Combining this with $BV\perp CA$ , we obtain BV || UO.

Now, if we consider not the triangle ABC with the points K and N on its sidelines AB and BC, but the triangle KBN with the points A and C on its sidelines KB and BN instead, then we similarly obtain BU || VO.

From BV || UO and BU || VO, it follows that the quadrilateral BUOV is a parallelogram. Hence, its diagonals BO and UV bisect each other, i. e. the midpoint R of the segment BO is also the midpoint of the segment UV.

The points B and M are the two common points of the circumcircles of triangles ABC and KBN; the centers of these two circumcircles are U and V. Thus, since the two common points of two circles are symmetric to each other with respect to the center line, it follows that the points B and M are symmetric to each other with respect to the line UV. Since the point R lies on the line UV, we thus get RB = RM. Hence, the point M lies on the circle with center R and radius RB. But since R is the midpoint of the segment BO, the circle with center R and radius RB is the circle with diameter BO; thus, the point M lies on the circle with diameter BO, so that < OMB = 90°, and the problem is solved.

Darij

This post has been edited 1 time. Last edited by darij grinberg, Mar 22, 2007, 5:48 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

armpist

527 posts

armpist

#3 h Nov 13, 2005, 8:38 PM • 5 Y

Y by Reynan, Melina_1535, Adventure10, wasikgcrushedbi, Mango247

Here is my solution:

Points B and M are the end-points of a common chord of two circles ABC and KBN.

Centers of these two circles are on the perpendicular bisector p of BM.

Circle OBM must also have it's center X on p. So X is the midpoint of OB. XM =XB = XO

makes angle BMO = 90

Thank you.

M.T.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

darij grinberg

6555 posts

darij grinberg

#4 h Nov 13, 2005, 9:05 PM • 2 Y

Y by Adventure10, Mango247

armpist wrote:

Here is my solution:

Points B and M are the end-points of a common chord of two circles ABC and KBN.

Centers of these two circles are on the perpendicular bisector p of BM.

Circle OBM must also have it's center X on p. So X is the midpoint of OB. XM =XB = XO

makes angle BMO = 90

Nice solution. Alas, 90% of the solution is missing...

Darij

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

armpist

527 posts

armpist

#5 h Nov 14, 2005, 5:57 AM • 5 Y

Y by Mathcat1234, Adventure10, Mango247, ehuseyinyigit, and 1 other user

But you managed to navigate over the gaps and voids, Darij Grinberg!

Consider yourself lucky, at least 10% is there. As you remember, J. Steiner

most of the times made statements without any proof whatsoever.

Also listen to this:

I hope that posterity will judge me kindly, not only as to the things which I have explained, but also to those which I have intentionally omitted so as to leave to others the pleasure of discovery. La Geometrie.

It was another old-timer Rene Descartes talking. Can you appreciate this?

I think I can, and hope you do too.

Thank you.

M.T.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Ashegh

858 posts

Ashegh

#6 h Apr 25, 2006, 5:09 PM • 3 Y

Y by Adventure10, SaintBroseph, Mango247

there exists nice solutions here.

hmmmm...

i dont see any solution with inversion.

i want to write it

.

problem:a circle with center $O$ passes through the vertices $B,C$ of triangle $ABC$ and intersects the segments

$AB,AC$ again at distinct points $F,S$ respectively.the circumcircles $ABC$ and $AFS$ ,intersects at exactly two

distinct points $M,A$ .prove that:

$\angle AMO=90$

ok...

lets invert the shape at a center $A$ and with the radius of power of $A$ to circle $(O)$ .

i have been said that i dont like to invertall the shape,because it may make it too busy.

then i just invert the important part of our shape.

before inversion:

first iwant to say that $AM,SF,BC$ are concurrent.suppose $AM, SF$ intersects at $M\"$.

$M\"$ is on the radical axis of circles $ABC,AFS$ .

it is also on radical axis of circles $(O),AFS$ .

it means that power of $M\"$ to circles $ABC,(O)$ is eaqual .and $M\",B,C$ should be colinear.

after inversion:

$S,F$ changes to $C,B$ respectively.and invert of M,should be colinear with them.

then $M\"$ should be the invert of $M$ .

suppose $K$ is a point inwhich $AK$ is tangent to circle $(O)$ .

the polar of $A$ to circle $(O)$ both goes through $K$ and $M\"$.then the polar is $KM\"$.

it is known that the polarof point $A$ to circle $(O)$ is prependicular to $AO$ .

name the intersection $AO,KM\"$ , $O'$ .

triangles $AKO'$ and $AKO$ , are simmilar to each other.then we have:

$AO'.AO=AK^2$ .

and it means that $O'$ is the inversive form of $O$ .

ok...

now after inversion we see that:

$\angle AO'M\"=90$

and it shows before inversion we should have: $\angle AMO=90$

and every thing is ok.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Andreas

578 posts

Andreas

#7 h Jan 26, 2007, 7:15 PM • 2 Y

Y by Adventure10, Mango247

Sorry to repost on this. I think I've a bit different solution.

Let $Q = CK \cap AN$ and $R = KN \cap AC$ . Then the line through $Q$ and $R$ is the polar of $B$ wrt circle with center $O$ .
Since $\angle BKC = \angle ANB = 90^{\circ}$ , then $Q$ belongs to the circumcircle of $BKN$ , and $BQ$ is a diameter.
Let $S = QR \cap BO$ , since $QR \bot BO$ then $S$ also belongs to the circumcircle.
Hence $BNSQKM$ is a hexagon inscribed in that circumcircle.
By the Radical Axis Theorem $MB$ intersect $AC$ in $R$ , and finally by Pascal's Theorem $M, Q, O$ are collinear, and we are done.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Virgil Nicula

7054 posts

Virgil Nicula

#8 h Jan 26, 2007, 9:40 PM • 4 Y

Y by iarnab_kundu, Adventure10, ehuseyinyigit, and 1 other user

Denote $L\in AC\cap BM$ and $S\in AN\cap CK$ . Prove easily that $L\in KN$ and the point $M$ is the Miquel's point of the complete quadrilateral $ACNKBL$ (the intersection between at least the circumcircles of the triangles $ABC$ and $BKN$ ). Because this quadrilateral is inscribed in the circle $C(O)$ it is well-known that the center $O$ is the orthocenter of the triangle $BLS$ (the quadrupoint $OBLS$ is orthocentrically). In conclusion, $OM\perp LB$ , i.e. $OM\perp MB$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Virgil Nicula

7054 posts

Virgil Nicula

#9 h Jan 27, 2007, 10:10 AM • 6 Y

Y by Adventure10, Mango247, Mango247, and 3 other users

An equivalent enunciation. Let $\triangle ABC$ and the points $M\in BC\ ,\ P\in AM$ .

The circumcircles $C(O_{b})$ , $C(O_{c})$ of the triangles $MPB$ , $MPC$ cut again

the lines $AB$ , $AC$ in the points $X$ , $Y$ respectively. Prove that

$\{\begin{array}{ccc}P\in BY & \Longleftrightarrow & O_{c}X\perp AB\\\ P\in CX & \Longleftrightarrow & O_{b}Y\perp AC\end{array}$ . What happen if $P\in CX\cap BY$ ?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Altheman

6194 posts

Altheman

#10 h Dec 18, 2007, 11:17 PM • 1 Y

Y by Adventure10

Let $AC\cap NK=M'$ , $AN\cap KC=X$ . $OX\cap BM'=M^*$ . $BO\cap XM'=H$ . Call the circumcircle of $\triangle AKN$ : $\Gamma$ . Let $\omega$ be the circle center $B$ that is orthogonal to $\Gamma$ .

Theorem: $XM'$ is the polar of $B$ wrt $\Gamma$ . $XB$ is the polar of $M'$ wrt $\Gamma$ .

Lemma 1: $OXM^*\perp BM'$ and $M'XH\perp BO$
Proof: Since $X$ is on the polar of $B$ and $M'$ , then $B$ , $M'$ are on the polar of $X$ so $BM'$ is the polar of $X$ . Now both facts follow since the line through a point and the center of the circle is perpendicular to that point's polar.

Lemma 2: Let $XM'\cap \Gamma=Y,Z$ . $\omega$ is the circle center $B$ with radius $BY$ .
Proof: Since $M'YXZ$ is the polar of $B$ , so $BY$ and $BZ$ are tangents to $\Gamma$ . But then the circle center $B$ , radius $BY=BZ$ is orthogonal to $\Gamma$ .

Lemma 3: $BM'*BM^*=r_\omega^2$
Proof: $\angle OHM'=\angle HM^*M'=90^\circ$ , so $OHM^*M'$ is cyclic. Bypower of a point, $BM'*BM^*=BH*BO$ . Since $\angle BYO=\angle YHB=90^\circ$ and a simple triangle similarity, $r_\omega^2=BY^2=BH*BO$ . The result follows.

Lemma 4: $r_\omega^2=BM*BM'$ .
Proof: Consider an inversion about $\omega$ . $\Gamma$ maps to itself. The circumcircle of $\triangle BNKM$ maps to line since $B$ is the center of inversion. Since $N$ maps to $C$ and $K$ maps to $A$ , the inversive image is $AC$ . Likewise, the circumcircle of $\triangle BCAM$ maps to $NK$ . Since $M$ is the intersection of these two circumcircles, the intersection of these two lines is the inversive image of $M$ . The result follows.

By lemmas 3 and 4, $BM=BM^*$ . But $M$ and $M^*$ are both on segment $BM'$ , so $M\equiv M^*$ , and the perpendicularity follows.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Altheman

6194 posts

Altheman

#11 h Dec 19, 2007, 8:57 AM • 5 Y

Y by Adventure10, Mango247, ehuseyinyigit, cubres, and 1 other user

Wow...you can pretty much angle chase this to death...

Let $AC$ , $NK$ , $BM$ intersect at $R$ , the radical center of the circles. $\angle NMR=\angle RKB=180-\angle AKN=180-\angle RCN$ , so $NCRM$ is cyclic.
Note if $\angle COK=2z$ , then $\angle CAK=\angle CMR=\angle CNR=\angle KNB=\angle KMB=z$ , so $\angle CMK=180-\angle KMB-\angle CMR=180-\angle COK$ , so $COKM$ is cyclic.

Let $\angle KMO=\angle OCK=y$ , then in triangle $OCK$ , $2z+y+y=180$ , so $y+z=90=\angle OMB$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

The QuattoMaster 6000

1184 posts

The QuattoMaster 6000

#12 h Mar 23, 2008, 5:36 PM • 4 Y

Y by JasperL, Adventure10, Mango247, ehuseyinyigit

orl wrote:

A circle with center $O$ passes through the vertices $A$ and $C$ of the triangle $ABC$ and intersects the segments $AB$ and $BC$ again at distinct points $K$ and $N$ respectively. Let $M$ be the point of intersection of the circumcircles of triangles $ABC$ and $KBN$ (apart from $B$ ). Prove that $\angle OMB = 90^{\circ}$ .

Here's a slightly different solution without use of the radial center:
Solution

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

The QuattoMaster 6000

1184 posts

The QuattoMaster 6000

#13 h Jul 6, 2008, 4:22 AM • 4 Y

Y by JasperL, ibadat, Adventure10, Elnuramrv

Sorry for reviving this topic again, but here is another neat solution to this problem:
Solution

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

simon89889

33 posts

simon89889

#14 h Aug 21, 2009, 7:25 AM • 1 Y

Y by Adventure10

armpist wrote:

Here is my solution:

Points B and M are the end-points of a common chord of two circles ABC and KBN.

Centers of these two circles are on the perpendicular bisector p of BM.

Circle OBM must also have it's center X on p. So X is the midpoint of OB. XM =XB = XO

makes angle BMO = 90

Thank you.

M.T.

I check this problem for many times
But the following answer is weird...
I think this answer has somthing wrong...
Or it isn't all right...
Could someone help me please??

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Agr_94_Math

881 posts

Agr_94_Math

#15 h May 1, 2010, 11:04 AM • 2 Y

Y by Adventure10, Mango247

Ashegh proved the problem with inversion of center $A$ .
But there is a simpler solution by using inversion with center $B$ .
Now consider an inversion with center $B$ and an arbitrary radius of inversion.
This would transform $A'. C', M'$ to be collinear and $K, N', M'$ also collinear in such a way that $A'C'N'K'$ forms a homothetic circle to $ACNK$ .
Now, to find the inverse of $O$ , consider the polar of $B$ with respect to $ACNK$ , that is $XY$ where $BX, BY$ are tangents from $B$ to the circle $ACNK$ .
Let after the inversion, the inverse images of $X,Y$ be $X', Y'$ . Now since $B, O, X, Y$ are concyclic, it implies $O', X', Y'$ are collinear . However, as $O$ was equidistant from $X,Y$ ; it implies that $O'$ is the midpoint of $X'Y'$ .
Now, $B$ is the pole of $KM$ . THis implies polar of $B$ passes through $M'$ . But $X'Y'$ is the polar of $B$ with respect to $A'C'N'K'$ .
Therefore, angles $OBM = BO'M' = BO'X' =BO'Y' = 90$ degrees. (As tangents from an exterior point are equal).

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

PCChess

548 posts

PCChess

#34 h Jun 26, 2021, 7:25 PM

Y by

Let $R=\overline{KC} \cap \overline{AN}$ and $P=\overline{KN} \cap \overline{AC}$ . Note that $M$ is the Miquel Point of $KNCA$ . It is well known that under inversion under $(KNCA)$ , $R$ inverts to $M$ . By Brocard's, it follows that $OM \perp BP$ , so $\angle OMB=90$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

HamstPan38825

8869 posts

HamstPan38825

#35 h Sep 20, 2021, 7:21 PM

Y by

$M$ is the Miquel Point of $ABCD$ . Then apply the Master Miquel Theorem.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Albert123

204 posts

Albert123

#37 h Nov 18, 2021, 5:10 AM

Y by

Let $KN \cap AC=L$ , $AN \cap KC=R$
Note that: $M$ is the Miquel - point of $ACNKBL$
But $AKNC$ is cyclic $\implies M \in LB$ .(This is since the radical axes of $\odot ABC, \odot AKNC, \odot BKN \implies BM, KN, AC$ are concurrents in $L$ ).
By chasing angles: $\angle BKM = \angle BNM = \angle MLC$
Let $\angle BMK=\angle BNK=\angle KAC=\angle NML=\alpha$
$\implies \angle KOR=\angle ROC=\alpha$
Lemma:
Let $ABCD$ be a cyclic quadrilateral, inscribed in a circle $\omega$ centered at $O$ .Let $AB \cap CD=P$ and $AD \cap BC=Q$ .
Let the intersection of the diagonals $AC$ and $BD$ be $R$ . Let $M$ be the Miquel point of $ABCDPQ$ . Then: $OM \perp PQ$
Proof:
By chasing angles: $M \in (BOD)$ . In a similar way, we can prove that $M \in (AOC)$ .
The line $AC$ is sent to the circle $(OAC)$ and the line $BD$ is sent to the circle $(OBD)$ .
The lines $AC$ and $BD$ intersect at $R \implies$ the image of $R=R^*$ is $R^* \in \odot AOC \cap \odot DOB=M$ .
$\implies O,R,M$ are collinears
By polars: $r \equiv PQ \implies OR \perp PQ \implies OM \perp PQ$
In the problem: $OM \perp LB \implies \angle OMB=90$ $\blacksquare$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

ike.chen

1162 posts

ike.chen

#38 h Nov 18, 2021, 5:22 AM

Y by

It's clear that $M$ is the Miquel Point of $ACNK$ . But $ACNK$ is also cyclic, so Master Miquel finishes. $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

math12345678

244 posts

math12345678

#39 h Nov 18, 2021, 5:25 AM

Y by

HamstPan38825 wrote:

$M$ is the Miquel Point of $ABCD$ . Then apply the Master Miquel Theorem.

ike.chen wrote:

It's clear that $M$ is the Miquel Point of $ACNK$ . But $ACNK$ is also cyclic, so Master Miquel finishes. $\blacksquare$

what is master miquel theorem?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

ike.chen

1162 posts

ike.chen

#40 h Nov 18, 2021, 5:32 AM

Y by

math12345678 wrote:

HamstPan38825 wrote:

$M$ is the Miquel Point of $ABCD$ . Then apply the Master Miquel Theorem.

ike.chen wrote:

It's clear that $M$ is the Miquel Point of $ACNK$ . But $ACNK$ is also cyclic, so Master Miquel finishes. $\blacksquare$

what is master miquel theorem?

This handout covers some the most essential properties of Complete Quadrilaterals.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Mogmog8

1080 posts

Mogmog8

#41 h Dec 20, 2021, 6:26 PM • 2 Y

Y by centslordm, ehuseyinyigit

Notice that $M$ is the Miquel point of cyclic quadrilateral $ACNK$ and $\overline{MB},\overline{KN},$ and $\overline{AC}$ concur by Radical Axis. We are done as $\overline{OM}\perp\overline{BL}$ where $L=\overline{KN}\cap\overline{AC}.$ $\square$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

PianoPlayer111

40 posts

PianoPlayer111

#42 h Feb 1, 2023, 5:44 PM

Y by

Let the intersection of $MK$ and $BC$ be $Q$ and let the intersection of $MO$ and $BC$ be $P$ . Can you help me by proving that $(C,Q;P,B)=-1$ ?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

lifeismathematics

1188 posts

lifeismathematics

#43 h Mar 20, 2023, 7:13 AM

Y by

We, first of all, join $MC$ and denote the intersection of $MC$ with circle having center $O$ as $L$

since points $A,K,L,C$ are concyclic we have $\angle{MLK}=A$ and also since points $M,B,C,A$ are concylic we have $\angle{BMC}=A$ now this gives $MB \parallel K$ .

so we just want to show that $OM \perp KL$

notice $OL=OK$ and if $\angle{BCM}=\angle{LKN}$ (angle subtended by chord $LN$ )
and hence $\angle{MBC}=180^{\circ}-(\angle{BCM}+A)$

and since $K,M,B,N$ are concyclic points we have $\angle{MKN}=\angle{BCM}+A$ or $\angle{LKN}+A$ which gives $\angle{MKL}=A$ which gives $ML=MK$ which gives quadrilateral $MKOL$ is a kite and hence $KL \perp OM$ and hence $OM \perp MB$ as desired $\blacksquare$

Attachments:

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Taco12

1757 posts

Taco12

#44 h Jun 25, 2023, 1:58 AM • 1 Y

Y by ehuseyinyigit

OTIS Version of IMO 1985/5 wrote:

Let $ABCD$ be a cyclic quadrilateral with center $O$ . Lines $AD$ and $BC$ meet at $P$ . The circumcircles of $PAB$ and $PDC$ meet at $M$ . Prove that $\angle OMP=90^{\circ}$ .

Note that $M$ is the Miquel point of $ABCD$ . Then, by Master Miquel, $\angle OMP=90^{\circ}$ as required.

This post has been edited 1 time. Last edited by Taco12, Jun 25, 2023, 1:59 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

shendrew7

799 posts

shendrew7

#45 h Dec 21, 2023, 7:24 PM

Y by

Define $AC \cap KN=D$ and $AN \cap CK=E$ . By Master Miquel, $K$ lies on $OE \cap BD$ , and by Brocard we have $OE \perp BD$ , giving us our result. $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Mathandski

773 posts

Mathandski

#46 h May 24, 2024, 11:23 PM

Y by

Light-weight angle chasing solution

Attachments:

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

AshAuktober

1013 posts

AshAuktober

#47 h Oct 9, 2024, 10:14 AM

Y by

As $M$ is the miquel point of $ACKN$ , the result is true by Master Miquel Theorem.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Eka01

204 posts

Eka01

#48 h Oct 10, 2024, 5:29 PM

Y by

Note that $M$ is the miquel point of the cyclic quadrilateral $ACNK$ so the result is pretty well known.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Ilikeminecraft

674 posts

Ilikeminecraft

#49 h Feb 16, 2025, 12:05 AM

Y by

$M$ is the miquel point of degenerate quad $ACBD.$ Then, by Master Miquel Theorem, $M$ and $Q$ are inverses, so $OMQ$ is collinear. By radical axis, $M, P, R$ are collinear. By Brocard's, $MPR\perp OQ.$

This post has been edited 2 times. Last edited by Ilikeminecraft, Feb 16, 2025, 12:18 AM

Z K Y