, 2019

We will prove that the point $T$ belongs to the circle $c_1(O, Γ, M)$ and the point $N$ belongs to the circle $c_2(O, A, Δ)$, that is the quadrilaterals $NAOΔ$ and $MOΓT$ are cyclic. Then we conclude the following: the line $AΔ$ is the radical axis of the circles $C$ and $c_2$, the line $ΓT$ is the radical axis of the circles $C$ and $c_1$ and the line $OΣ$ is the radical axis $C_1$ and $C_2$, and therefore the lines $AΔ$, $ΓT$ and $OΣ$ are passing through the radical center of the three circles.

Since $BΔ$ is diameter of the circle $c$, we have $B\hat{A}Δ = 90^\circ$ and so $N\hat{A}Δ = 180^\circ - B\hat{A}Δ = 90^\circ$. Also, $ON$ is the perpendicular bisector of the segment $BΔ$, and so: $N\hat{O}Δ = 90^\circ$. From the last two angle equalities we conclude that the quadrilateral NAOΔ is cyclic.

The triangle $NBΔ$ is isosceles with $NB = NΔ$ (because $N$ belongs to the perpendicular bisector $BΔ$), and so $N\hat{B}Δ = N\hat{A}B$. Therefore taking in mind that $OA = OB = OΔ = OT = R$ we conclude that the isosceles triangles $OAB$ and $OΔT$ are equal. Therefore
$$(\alpha) \ OA = OT \text{ and } (\beta) \ AB = ΔT \Rightarrow NA = NT.$$
Hence the triangles $OAN$ and $OTN$ are equal, because they have their sides equal one to one and so
$$N\hat{O}A = N\hat{O}T = \frac{1}{2} \cdot A\hat{O}T, \qquad (1)$$
Moreover we have that
$$M\hat{Γ}T = A\hat{Γ}T = \frac{1}{2} \cdot A\hat{O}T \qquad (2)$$
From relations (1) and (2) it follows that $M\hat{Γ}T = N\hat{O}T = M\hat{O}T$, from which we conclude that the quadrilateral MOΓT is cyclic.