Let $B'\equiv{D}$, $C'\equiv{E}$, $B''\equiv{K}$ for convenience. Let ray $DE$ intersect the circumcircle at $L$. We will prove that $AK=AL$. Note that $\angle{AKL}=\pi-\angle{ADK}-\angle{KAC}=\angle{B}-\angle{KBC}$ and $\angle{ALK}=\angle{ALC}-\angle{KLC}=\angle{B}-\angle{KBC}$, hence $\angle{AKL}=\angle{ALK}=a_A \Longleftrightarrow AK=AL$. From the law of sines, we get: $$\frac{AK}{\sin{a_A}}=\frac{AB}{\sin{\angle{C}}}$$$$and$$$$\frac{AK}{\sin{(\pi-\angle{B})}}=\frac{AD}{\sin{a_A}}$$Dividing the two relationships gives: $$\frac{\sin{\angle{B}}}{\sin{a_A}}=\frac{AB \cdot \sin{a_A}}{AD \cdot \sin{\angle{C}}} \Longleftrightarrow$$$$\sin^2{a_A}=\frac{\sin{\angle{B}} \cdot \sin{\angle{C}} \cdot AD}{AB} \Longleftrightarrow$$$$\sin^2{a_A}=\sin{\angle{B}} \cdot \sin{\angle{C}} \cdot \cos{\angle{A}}$$In a similar manner, we obtain the relationships: $$\sin^2{a_B}=\sin{\angle{C}} \cdot \sin{\angle{A}} \cdot \cos{\angle{B}}$$$$\sin^2a_C=\sin{\angle{A}} \cdot \sin{\angle{B}} \cdot \cos{\angle{C}}$$Multiplying these, we get: $$\sin{a_A} \cdot \sin{a_B} \cdot \sin{a_C}=\sin{\angle{A}} \cdot \sin{\angle{B}} \cdot \sin{\angle{C}} \cdot \sqrt{\cos{\angle{A}} \cdot \cos{\angle{B}} \cdot \cos{\angle{C}}}$$Now, from $AM-GM$, we obtain: $$\sin{\angle{A}} \cdot \sin{\angle{B}} \cdot \sin{\angle{C}} \le \frac{(\sin{\angle{A}}+\sin{\angle{B}}+\sin{\angle{C}})^3}{27}$$The second derivative of the function $\sin{x}$ is $-\sin{x}<0$, so $\sin{x}$ is concave in $(0, \frac{\pi}{2})$. From Jensen's inequality, we have: $$\sin{\angle{A}}+\sin{\angle{B}}+\sin{\angle{C}} \le 3 \cdot \sin{\frac{\angle{A}+\angle{B}+\angle{C}}{3}}=3\sin{\frac{\pi}{3}}=\frac{3\sqrt{3}}{2}$$Hence: $$\sin{\angle{A}} \cdot \sin{\angle{B}} \cdot \sin{\angle{C}} \le \frac{(\frac{3\sqrt{3}}{2})^3}{27}=\frac{3\sqrt{3}}{8}$$Similarly, we observe that the second derivative of $\cos{x}$ is $-\cos{x}<0$ in $(0, \frac{\pi}{2})$, so: $$\cos{\angle{A}} \cdot \cos{\angle{B}} \cdot \cos{\angle{C}} \le \frac{(\cos{\angle{A}}+\cos{\angle{B}}+\cos{\angle{C}})^3}{27} \le \frac{(3\cos{\frac{\angle{A}+\angle{B}+\angle{C}}{3}})^3}{27}=\cos^3{\frac{\pi}{3}}=\frac{1}{8} \Longleftrightarrow$$$$\sqrt{\cos{\angle{A}} \cdot \cos{\angle{B}} \cdot \cos{\angle{C}}} \le \frac{1}{2\sqrt{2}}=\frac{\sqrt{2}}{4}$$In conclusion, we have: $$\sin{a_A} \cdot \sin{a_B} \cdot \sin{a_C} \le \frac{3\sqrt{3}}{8} \cdot \frac{\sqrt{2}}{4}=\frac{3\sqrt{6}}{32}$$Equality holds when $\angle{A}=\angle{B}=\angle{C}=\frac{\pi}{3}$.

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Turns out one can solve this without introducing new points (just Sine Law on appropriate triangles).