# Property of bicentric quadrilateral involving the Miquel point of $OMXN$

Let $$ABCD$$ be a bicentric quadrilateral, let $$O$$ be its circumcenter and $$I$$, its incenter. Let $$M$$ and $$N$$ be the midpoints of $$AC$$ and $$BD$$ repectively. Let $$X$$ be the intersection of $$AC$$ and $$BD$$. Let $$P$$ be the second intersection of circles (IAC) and (IBD). Prove that P is the Miquel point of the quadrilateral $$OMXN$$.

My work:

I have managed to prove that $$P$$,$$X$$,$$I$$ and $$O$$ are colinear using properties of the radical axis $$IP$$. We also know that $$M$$,$$N$$ and $$I$$ are colinear because of the Newton line of a circumscriptible quardrilateral. So $$I$$ is actually the intersection point of the diagonals of our quadrialteral $$OMXN$$. We also know that the quadrilateral $$OMXN$$ is cyclic and the diameter of its circumcircle is $$OX$$ so we know that its Miquel point is located on $$FG$$ where {$$F$$}=$$OM\cap XN$$ and {$$G$$}=$$ON\cap XM$$. Let $$M_q$$ be the Miquel point of $$OMXN$$. Because $$FM\perp MX$$ and $$GN\perp NX$$ we know that $$X$$ is the orthocenter of $$\triangle OFG$$ so $$OX\perp FG$$ but $$XM_q\perp FG$$ (because $$XNGM_q$$ is cyclic by definition and $$XN\perp NG$$) so $$O$$,$$X$$ and $$M_q$$ are collinear. Now we can define $$M_q$$ as the intersection of $$OX$$ and $$FG$$ so if we prove that $$P$$ also lies on $$FG$$ we are done

Edit: From Brocard's theorem we know that $$FG$$ is the polar line of $$I$$ with respect to the circumcircle of $$OMXN$$ so we only need to prove that $$P$$ lies on this polar line which translates to proving that $$(O,X;I,P)$$ is a harmonic division which seems way easier to prove than the initial statement but I think that it is still pretty tough.

Source:

This is actually something I have dicovered while playing with bicentric quadrilaterals on Geogebra. It seems like it is true so I hope it is lol. However I am quite sure that it is true and I am curious why is that so

• I have proved it, it is a really hard problem... Jan 29, 2022 at 23:57

First, we will proof some lemma:

Lemma 1: Suppose a line $$l$$ pass $$A,B,C,D$$ four points, let $$B',C',D'$$ be the image when doing inversion of $$A$$. Then $$(A,C;B,D)$$ is a harmonic division if and only if $$C'$$ is the midpoint of $$B'D'$$.

Proof: Let $$A$$ be zero, and $$B,C,D$$ has the coordinate $$b,c,d$$. So we have $$(A,C;B,D)$$ is a harmonic division if and only if $$(c-b)d=b(d-c)$$, if and only if $$1/b+1/d=2/c$$, if and only if $$C'$$ is the midpoint of $$B'D'$$.

Then we use inversion to finish your proof:

We use inversion w.r.t. the incircle of $$ABCD$$. Let four tangent points of $$AD,DC,CB,BA$$ to be $$Q,R,S,T$$. Let $$E,F,G,H$$ be the midpoint of $$QT,TS,SR,RQ$$. So $$E,F,G,H$$ are images of $$A,B,C,D$$ respectively. Furthermore, let $$K$$ be the intersection of $$EG$$ and $$HF$$, so $$K$$ is the image of $$P$$. Let $$L$$ to be the image of $$O$$. Let the circumcircle of $$HIF$$ and $$EIG$$ be $$J$$. So we only need to proof, by lemma 1, $$JK=KL$$. Now we seek for another description of $$L$$. Notice that $$O$$ is the intersection of the perpendicular bisector of $$BD$$ and the perpendicular bisector of $$AC$$. Let $$HIFU$$ be the harmonic quadrilateral, so $$L$$ is on the circle passing $$I$$, $$U$$, and orthogonal to the circumcircle of $$HIF$$. So we know that $$\angle ULI=90^\circ-\angle UJI$$. Similarly, let $$V$$ be the point such that $$EIGV$$ is the harmonic quadrilateral, so $$\angle VLI+\angle VGI=90^\circ$$.

Notice that $$I,J,K,L$$ are collinear* (this is because $$O,X,I,P$$ collinear), and $$K$$ is the intersection of $$IK$$ and circumcircle of $$FIH$$. So $$JU\parallel HF$$. Similarly, $$VJ\parallel EG$$. Since $$K$$ is the midpoint of $$EG$$ and $$VJ\parallel EG$$, we have $$KJ=KV$$. Similarly, $$KJ=KU$$. So $$K$$ is the circumcenter of $$UJV$$. Since $$I,J,L$$ are collinear, and also $$\angle VLI+\angle VGI=90^\circ$$ and $$\angle ULI=90^\circ-\angle UJI$$, we have $$\angle JUL=\angle JVL=90^\circ$$. So $$J,V,L,U$$ are on the same circle, and since $$\angle JVL=90^\circ$$, $$JL$$ is the diameter. Since $$K$$ is the circumcenter of $$UJV$$, we have $$KJ=KL$$. So the claim is proved.

• If we don't know this is collinear, we can first using radical axis of circumcircles of $$HIF$$, $$EIG$$ and $$EFGH$$ to proof $$I,J,K$$ collinear, then we can proof that $$U,I,V$$ collinear (this can be derived from $$\angle EIH+\angle HIG=180^\circ$$). Then proof $$K$$ to be the circumcircle of $$JUV$$. Then, we let $$L'$$ to be the intersection of the perpendicular line of $$JU$$ passing $$U$$ and the perpendicular line of $$JV$$ passing $$V$$. We have $$J,U,L',V$$ on the same circle and $$I,J,K,L'$$ are collinear. Then we have $$\angle UL'K=90^\circ-\angle UJK=\angle ULK$$ and $$\angle VL'K=90^\circ-\angle VJK=\angle VLK$$, so $$L=L'$$. So $$I,J,K,L$$ are collinear,
• Wow, nice proof. Honestly, I wasn't expecting it to be this hard but I think that it is a pretty nice property with a cool proof Jan 30, 2022 at 13:01