* Question* : Do you know this property of a hexagon?

Consider the configuration: Six points $A_1$, $A_2$, $A_3$, $A_4$, $A_5$, $A_6$ in a plane and let six points $B_i \in A_iA_{i+1}$ for $i=1, 2,\dots, 6$ and $7\equiv 1$.

- Let six points $B_1$, $\cdots$, $B_6$ are collinear and $$\frac{\overline{B_1A_1}}{\overline{B_1A_2}}. \frac{\overline{B_2A_2}}{\overline{B_2A_3}}. \frac{\overline{B_3A_3}}{\overline{B_3A_4}}.\frac{\overline{B_4A_4}}{\overline{B_4A_5}}. \frac{\overline{B_5A_5}}{\overline{B_5A_6}}. \frac{\overline{B_6A_6}}{\overline{B_6A_1}}=1$$ then six points $A_1$, $A_2$, $A_3$, $A_4$, $A_5$, $A_6$ lie on a conic.
- Let six points $A_1$, $A_2$, $A_3$, $A_4$, $A_5$, $A_6$ lie on a conic and five points $B_1$, $\cdots$, $B_5$ lie on line $(L)$ and $$\frac{\overline{B_1A_1}}{\overline{B_1A_2}}. \frac{\overline{B_2A_2}}{\overline{B_2A_3}}. \frac{\overline{B_3A_3}}{\overline{B_3A_4}}.\frac{\overline{B_4A_4}}{\overline{B_4A_5}}. \frac{\overline{B_5A_5}}{\overline{B_5A_6}}. \frac{\overline{B_6A_6}}{\overline{B_6A_1}}=1$$ then $B_6$ also lie on $(L)$

**See also**:

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