, 2011

Let $\gamma$ be the circle through the $A_i$, centred at $O$, and let tangents to $\gamma$ at $A_i$ and $A_{i+1}$ meet at $B'_i$. Notice that the line $B'_iB'_{i+3}$ is the image of the circle $B_iOB_{i+3}$ under the inversion of pole $O$ and power $r^2$, where $r$ is the radius of $\gamma$. By Brianchon's theorem, the three lines $B'_iB'_{i+3}$ are concurrent at a point $Q$. Notice further that $Q$ is different from $O$, for no opposite sides of the hexagon $A_0A_1\cdots A_5$ are parallel. Consequently, the three circles $B_iOB_{i+3}$ share a second point $P$, different from $O$: the image of the point $Q$ under the inversion.

The lemma below shows that the points $P, C_i, C_j$ and $D_k$ are concyclic (not necessarily in this order), and the lines $PO$ and $PD_k$ are isogonal with respect to the lines $PC_i$ and $PC_j$, where $\{i, j, k\} = \{0, 1, 2\}$. Finally, a standard angle-chase argument shows that $P$ and the three points $D_i$ are concyclic: with reference to the figure below, write successively

$$\begin{aligned} \angle D_i P D_j &= \angle D_i P C_i + \angle C_i P D_j \\ &= \angle D_i P C_i + \angle C_k P O \quad (\text{for } P D_j \text{ and } P O \text{ are isogonal relative to } P C_k \text{ and } P C_i) \\ &= \angle D_i P C_i + \angle C_j P D_i \quad (\text{for } P C_k \text{ and } P C_j \text{ are isogonal relative to } P D_i \text{ and } P O) \\ &= \angle C_i P C_j \\ &= \angle C_i D_k C_j \quad (\text{for } P, C_i, C_j, D_k \text{ are concyclic}) \\ &= \angle D_j D_k D_i, \end{aligned}$$

to conclude that the points $P, D_0, D_1$ and $D_2$ are indeed concyclic.

Lemma. Two circles, $\gamma_1$ and $\gamma_2$, meet at the points $X$ and $Y$. A line through $Y$ meets again $\gamma_1$ at the point $Y_1$, and $\gamma_2$ at the point $Y_2$. The tangent to $\gamma_1$ at $Y_1$ meets the tangent to $\gamma_2$ at $Y_2$ at the point $Z$. Then the points $X, Z, Y_1$ and $Y_2$ are concyclic, and the lines $XY$ and $XZ$ are isogonal with respect to the lines $XY_1$ and $XY_2$.

Proof. If $X$ and $Z$ lie on opposite sides of the line through $Y$, then the angle $Y_1 X Y_2$ is the sum of the angles $X Y_1 Y$ and $X Y_2 Y$ which are respectively equal to the angles $Y_1 Z Y_2$ and $Y_2 Z Y_1$, whose sum is supplementary to the angle $Y_1 Z Y_2$. Consequently, the quadrangle $X Y_1 Z Y_2$ is cyclic. It then follows that the angles $Z X Y_2$ and $Z Y_1 Y_2$ are equal, and since the latter is equal to the angle $X Y_1 Y$, we conclude that the lines $XY$ and $XZ$ are indeed isogonal with respect to the lines $XY_1$ and $XY_2$.
If $X$ and $Z$ lie on the same side of the line through $Y$, then the angle $Y_1 X Y_2$ is the difference of the angles $X Y_1 Y$ and $X Y_2 Y$ in some order, depending on which side of the line $XY$ the line $Y_1 Y_2$ is situated. The later angles are respectively supplementary to the angles $Y_1 Z Y$ and $Y_2 Z Y$ whose difference in the corresponding order is equal to the angle $Y_1 Z Y_2$. Consequently, the quadrangle $X Y_1 Y_2 Z$ or $X Y_2 Y_1 Z$ is cyclic. Isogonality is proved by adapting the argument in the former case.