Miquel Point of Cyclic Quad

Question

Can anyone help me prove the following property?

Let $ABCD$ be a cyclic quadrilateral with circumcircle $w$ and with no two sides parallel. Let its diagonals meet at $N$ and let the Miquel point of this quadrilateral be $M$. Then $M$ is the inverse of $N$ with respect to inversion in the circumcircle $w$.

Apparently this is shown through angle chasing, but I am not able to do it.

Answer 1

Yeah, if you define it like on the picture (Clifford's configuration), then it is simple. Let $\angle \, F = \angle \, CFD = \phi$ and $\angle \, E = \angle \, AED = \epsilon$ and $\angle \, B = \angle \, ABC = \beta$. Let $O$ be the center of the circumcircle of $ABCD$.

1. $\angle \, AOC = 2 \angle \, ABC = 2 \beta$.

2. $\angle \, CMD = \angle \, CFD = \phi$ and $\angle \, AMD = \angle \, AED = \epsilon$. Hence $\angle \, AMC = \angle \, AMD + \angle \, CMD = \epsilon + \phi.$

3. $\angle \, ADE = \angle \, CDF = \angle \, ABC = \beta$ because $ABCD$ is cyclic.

4. $\angle \, BAD = \angle \, AED + \angle \, ADE = \epsilon + \beta$. Furthermore, $ \angle \, BCD = \angle \, CFD + \angle \, CDF = \phi + \beta$.

5. $\angle \, BAD + \angle \, BCD = \epsilon + \beta + \phi + \beta = 2 \beta + \epsilon + \phi = \pi$ because $ABCD$ is cyclic.

6. $\angle \, AOC + \angle \, AMC = 2 \beta + \epsilon + \phi = \pi$ which means that $AMCO$ is inscribed in a circle $c_F$.

7. By analogous angle chasing involving the bigger circles $BCME$ and $BAMF$ (plus Clifford's theorem that all four solid-line circles pass through $M$) one can show that $DMBO$ is also inscribed in a circle $c_E$.

8. Both circles $c_E$ and $c_F$ pass through $M$ by construction. If one performs an inversion with respect to the circumcircle $w$ of $ABCD$, the diagonal $AC$ maps to $c_F$ and $BD$ to $c_E$. Hence the intersection point $N$ of $AC$ and $BD$ is mapped to the intersection point (different from the center $O$) of $c_F$ and $c_E$, which is $M$.