Let $\alpha, \beta, \gamma$ be the angles $\widehat{O A B}, \widehat{O C D}$, and $\widehat{O E F}$. It is easy to see that $\alpha+\beta+\gamma=180^{\circ}$.

We will use the following result: if $O, X, Y, Z$ are four points such that $O X=O Y=O Z$ and $\widehat{X O Y}+\widehat{Y O Z}=180^{\circ}$, then the triangles $X O Y$ and $Y O Z$ have the same area.

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Indeed, by taking the sides $[O X]$ and $[O Z]$ as bases, we observe that these triangles have bases of the same length and their heights from $Y$ coincide.

We deduce that the areas of $O A B, O C D$, and $O E F$ are respectively equal to those of $O D F, O F B$, and $O B D$. By adding these areas, we obtain that half the area of the hexagon is equal to the area of $B D F$.