# Six points in a full circular quadrilateral belong to one circle

Today while using the Geogebra program I discovered a new engineering feature, I don't know if it was already known or not

Data: $$R$$ is a circle with center $$O$$, $$A,B,C,D∈R$$, $$AD∩BC=E$$, $$AD∩BC=F$$, $$R_1$$ is the passing circle passing through the points $$A,B,E$$, its center is $$M$$, and $$R_2$$ is the passing circle passing through the points $$A,C,D$$, its center is $$N$$, $$R_3$$ is the passing circle passing through the points $$B,C,F$$, its center is $$P$$, and $$R_4$$ is the passing circle passing through the points $$A,D,F$$, its center is $$Q$$, and $$R_1∩R_2∩R_3∩R_4=S$$. Required: Prove that the points $$M,N,O,P,Q,S$$ lie on one circle.

But I don't know how to prove it.

If the feature is already known, please mention a source that talks about it, and thank you

• Interesting. First hint I can suggest is to get rid of R3 and R4 as if you prove it for R1 and R2 it will be straightforward then to finish with the circles 3 and 4 Commented May 7 at 18:58
• It seems that five of them are known to lie on one circle, the rest of the proof that the center of the circle R belongs to the same circle artofproblemsolving.com/wiki/index.php/Miquel%27s_point Commented May 7 at 19:08
• Are you sure there is no additional condition in your case? In the link you gave it seems the center of the initial triangle (called O in your case) would not be cocyclic with the other centers. Maybe BCD isocele? Commented May 7 at 19:39
• Fun Fact: $S$ is the foot of the perpendicular from $O$ to line $EF$. Moreover, this perpendicular passes through $AC \cap BD$.
– Blue
Commented May 7 at 19:50
• The link i provided talks about 4 points in the general case, while here it talks about 4 points that lie on a circle @Martigan Commented May 8 at 0:02

From a point outside the given circle $$R$$ we draw two lines that determine the points $$A,B,C,D$$ of the circle which in turn determines the point $$E$$.
The center $$Q$$ of $$R_4$$ is the intersection of the bisectors of the sides $$AF$$ and $$FD$$ of the triangle $$\triangle{AFD}$$ and the center $$P$$ of $$R_3$$ is the intersection of the bisectors of the sides $$BF$$ and $$FC$$ of the triangle $$\triangle{BFC}$$.
The points $$O,P,Q$$ determine an unique circle (which will be the red circle of the statement).
A procedure exactly the same as the one we have just given determines the centers $$M$$ of $$R_1$$ and $$N$$ of $$R_2$$ and then we can optionally verify that $$M$$ and $$N$$ belong to the red circle or see that the equation of the circumcircle of the triangle $$\triangle{OMN}$$ coincides with the preceding one of the triangle $$\triangle{OPQ}$$.