#8 Complete Quadrilaterals and Miquel Points

By the converse of the inscribed angle criterion, equal angles subtending the same segment \(AB\) imply that the vertices of these angles lie on one circle with \(A\) and \(B\). Hence \(A,B,X,Y\) are concyclic.

Let \(A=l_1\cap l_2\), \(B=l_2\cap l_3\), \(C=l_3\cap l_4\), \(D=l_4\cap l_1\), \(E=l_1\cap l_3\), \(F=l_2\cap l_4\). The four triangles are \(ABE\), \(BCF\), \(CDE\), \(ADF\).

Inscribed angles subtending the same chord \(BE\) are equal. Therefore \(\angle BME=\angle BAE\).

Angles \(\angle BMF\) and \(\angle BCF\) subtend the same segment \(BF\). Their equality means that points \(M\) and \(C\) lie on one circle with \(B\) and \(F\). Hence \(B,C,F,M\) are concyclic.

They are \((ABE)\), \((ADF)\), \((CDE)\), and \((BCF)\). By Miquel's theorem, they have one common point.

The triples of these lines give circles \((ABF)\), \((BCE)\), \((CDF)\), and \((DAE)\). They pass through the Miquel point of the complete quadrilateral of the side lines.

We have \(\angle BMF=\angle BMA+\angle AMF\). From \(A,B,E,M\) cyclic, \(\angle BMA=\angle BEA\), the angle between \(l_3\) and \(l_1\). From \(A,D,F,M\) cyclic, \(\angle AMF=\angle ADF\), the angle between \(l_1\) and \(l_4\). The sum is the angle between \(l_3\) and \(l_4\), namely \(\angle BCF\). Therefore \(B,C,F,M\) are cyclic.

From the previous problem we already know that \(M\in (BCF)\). Hence \(\angle CMB=\angle CFB\), which is the angle between lines \(l_4\) and \(l_2\). Also, from \(M\in (ABE)\) we get \(\angle BME=\angle BAE\), the angle between \(l_2\) and \(l_1\). Therefore \(\angle CME=\angle CMB+\angle BME\) is the angle between \(l_4\) and \(l_1\), that is \(\angle CDE\). By the converse cyclicity criterion, \(C,D,E,M\) lie on one circle.

Let \(M\) be the second intersection of \((ABE)\) and \((ADF)\). By Problems 7 and 8, point \(M\) also lies on \((BCF)\) and \((CDE)\). Therefore all four circles pass through \(M\).

The four lines \(AB,BC,CD,DA\) form a complete quadrilateral. Its four circles are exactly \((ABF)\), \((BCE)\), \((CDF)\), and \((DAE)\). By Miquel's theorem, they pass through one point.

By the definition of the Miquel point, \(M\in (BCF)\). Thus \(B,C,F,M\) lie on one circle, and angles \(\angle BMF\) and \(\angle BCF\) subtend chord \(BF\). Therefore they are equal.

Since \(A,B,E,M\) are cyclic, \(\angle BME=\angle BAE\). Since \(A,D,F,M\) are cyclic, \(\angle DMF=\angle DAF\). Angles \(\angle BAE\) and \(\angle DAF\) are the same angle between lines \(l_2\) and \(l_1\). Hence \(\angle BME=\angle DMF\).

If the sum of opposite angles of quadrilateral \(AXBY\) is \(180^\circ\), then the quadrilateral is cyclic. Therefore points \(A,X,B,Y\) lie on one circle.

Lines \(AB,BC,CD,DA\) form a complete quadrilateral. Two of its circles, \((ABF)\) and \((BCE)\), meet again at \(M\). By Miquel's theorem, the same point lies on the other two circles of the complete quadrilateral: \((CDF)\) and \((DAE)\).

By Miquel's theorem, point \(M_1\), constructed as the second intersection of \((ABE)\) and \((ADF)\), also lies on \((BCF)\) and \((CDE)\). Therefore it is the common point of these two circles different from their already known common point \(C\). Hence it is \(M_2\).

Angles \(\angle BXF\) and \(\angle BMF\) subtend segment \(BF\). By the converse of the inscribed angle criterion, points \(X\) and \(M\) lie on one circle with \(B\) and \(F\). Hence \(B,F,M,X\) are concyclic.

Since \(M\in (BCF)\), angles \(\angle BMF\) and \(\angle BCF\) subtend chord \(BF\), so they are equal. Since \(M\in (DAE)\), angles \(\angle DME\) and \(\angle DAE\) subtend chord \(DE\), so they are also equal.

Since \(A,B,E,M\) are cyclic, \(\angle BME=\angle BAE\). Since \(A,D,F,M\) are cyclic, \(\angle DMF=\angle DAF\). Both right-hand angles are the angle between lines \(l_2\) and \(l_1\), so \(\angle BME=\angle DMF\).

For example, construct circles \((ABE)\) and \((ADF)\). They share point \(A\). Their second intersection \(M\) is the Miquel point. By Miquel's theorem, this point automatically lies on \((BCF)\) and \((CDE)\).

From \(A,B,E,M\) cyclic, \(\angle BMA=\angle BEA\). From \(A,D,F,M\) cyclic, \(\angle AMF=\angle ADF\). The sum of these angles equals the angle between \(l_3\) and \(l_4\), namely \(\angle BCF\). Therefore \(\angle BMF=\angle BCF\), hence \(B,C,F,M\) are cyclic.

Since \(M\) lies on \((ABE)\) and \((ADF)\), the angle proof of Miquel's theorem implies that \(M\) must lie on \((CDE)\). Directly, \(\angle CME\) splits into a sum of angles equal to \(\angle CDE\). By the converse of the inscribed angle criterion, \(C,D,E,M\) are cyclic.

The lines \(AB,BC,CD,DA\) form a complete quadrilateral. Since \(M\) lies on circles \((ABF)\) and \((BCE)\), Miquel's theorem gives \(M\in (CDF)\). In circle \((CDFM)\), the angles \(\angle DMC\) and \(\angle DFC\) subtend the same chord \(DC\), so they are equal.

Since \(A,B,E,M\) are concyclic, \(\angle BME=\angle BAE\). Since \(A,D,F,M\) are concyclic, \(\angle DMF=\angle DAF\). The angles \(\angle BAE\) and \(\angle DAF\) are the same angle between the lines \(l_2\) and \(l_1\), so \(\angle BME=\angle DMF\). Also, \(C,D,E,M\) are concyclic; the angles \(\angle CME\) and \(\angle CDE\) subtend chord \(CE\), hence they are equal.

Label the intersection points as \(A,B,C,D,E,F\). The four circles necessarily have the form \((ABE)\), \((ADF)\), \((BCF)\), \((CDE)\). If three of them pass through \(M\), then by choosing two circles with a common vertex, we get \(M\) as the Miquel point of the complete quadrilateral. By Miquel's theorem, \(M\) also lies on the fourth circle.