Truly very innovative! Call the center of the semi-circle ${O}$, the midpoints of $A_2A_8, A_3A_7$ be $M_2, M_3$ and ${Z}$'s pedal to them be $H_2, H_3$. Call the area of polygon ${P}$ $S_P$. It is well known (by harmonic points) that $A_2A_8, A_3A_7, XY$ are concurrent at the invertion of ${Z}$ wrt the semicircle, and that $A_2A_8ZO, A_3A_7ZO$ are respectively concyclic. So $\triangle A_2M_2O, \triangle A_8M_2O , \triangle A_3M_3O , \triangle A_8M_3O$ are all congurent triangles of $90^{\circ}-54^{\circ}-36^{\circ}$, so $S_ {\triangle A_2A_8O}=S_ {\triangle A_3A_7O}$. Since $\frac{ZH_2}{ZH_3}=\frac{OM_2}{OM_3}$, $S_ {\triangle A_2A_8Z}=S_ {\triangle A_3A_7Z}\ (*)$, using $S_{A_2A_3A_7A_8OZ}$ to subtract both sides of $(*)$ gets the result in the problem.

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Note that $[A_2A_3A_7A_8]=[ZA_2A_3]+[ZA_3A_7]+[ZA_7A_8]-[ZA_2A_8]$, so it suffices to show $[ZA_3A_7]=[ZA_2A_8]$. $[ZA_3A_7]=\frac 12 \cdot ZA_3\cdot ZA_7\cdot \sin (\angle A_7ZA_3)$ and $[ZA_2A_8]=\frac 12 \cdot ZA_2\cdot ZA_8\cdot \sin (\angle A_8ZA_2)$. Since $\angle A_7ZA_3=72^\circ$ and $\angle A_8ZA_2=108^\circ$, it suffices to show that $ZA_7\cdot ZA_3=ZA_2\cdot ZA_8$. Let $A_7',A_8'$ be the reflections of $A_7,A_8$ over $XY$. Then $A_2-Z-A_8'$, $A_3-Z-A_7'$, and $A_2,A_3,A_7',A_8'$ are on $(XY)$. Thus $ZA_7\cdot ZA_3=ZA_7'\cdot ZA_3=ZA_2\cdot ZA_8'=ZA_2\cdot ZA_8$, as desired.

In fact, we only need to prove that $\triangle A_2A_7O$ is similar to $\triangle A_3A_8O$ Notice that $\triangle A'_3A'_8O$ is equivalent to $\triangle A_2A_7O$ (A' represents the other intersection point of AZ with $circle XY$ ) So we are done

Notice that $$[A_2A_3A_7A_8]=[ZA_2A_3]+[ZA_7A_8]+[ZA_3A_7]-[ZA_2A_8]$$So it suffices to show that $[ZA_3A_7]=[ZA_2A_8]$. Reflect $A_7$ and $A_8$ across $XY$ to $B_7$ and $B_8$. As $\sin(\angle A_2ZA_8)=$ $\sin(\angle A_3ZA_7)$ and $ZA_3\cdot ZA_7=$ $ZA_3\cdot ZB_7= ZA_2\cdot ZB_8=ZA_2\cdot ZA_8$, we are done by the sine area formula.

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