How can prove this equation. if  $a+b=c+d=e+f=\dfrac{\pi}{3}$,
$\dfrac{\sin{a}}{\sin{b}}\cdot\dfrac{\sin{c}}{\sin{d}}\cdot\dfrac{\sin{e}}{\sin{f}}=1$,
Prove that:
$\dfrac{\sin{(2a+f)}}{\sin{(2f+a)}}\cdot\dfrac{\sin{(2e+d)}}{\sin{(2d+e)}}\cdot\dfrac{\sin{(2c+b)}}{\sin{(2b+c)}}=1$
 A: Consider an equilateral triangle $ABC$, and let $D$ be on $BC$ so that $\angle{BAD}=a$, so $\angle{DAC}=\frac{\pi}{3}-a=b$. Let $E$ be on $AC$ so that $\angle{CBE}=c$, so $\angle{EBA}=\frac{\pi}{3}-c=d$. Let $F$ be on $AB$ so that $\angle{ACF}=e$, so $\angle{FCB}=\frac{\pi}{3}-e=f$.By the sine version of Ceva's theorem and the given condition $\frac{\sin{a}}{\sin{b}}\cdot \frac{\sin{c}}{\sin{d}}\cdot \frac{\sin{e}}{\sin{f}}=1$, $AD, BE, CF$ are concurrent at a point, which we shall call $P$. 
Extend $AD$ to points $A_1, A_2$ s.t. $\angle{A_1CB}=a+f, \angle{A_2BC}=b+c$. We have $\angle{CA_1A}=\pi-\angle{A_1CA}-\angle{A_1AC}=\pi-b-(e+f+a+f)=e$. Similarly $\angle{BA_2A}=d$. Thus $APC$ is similar to $ACA_1$ and $APB$ is similar to triangle $ABA_2$. Therefore $\frac{AA_1}{AC}=\frac{AC}{AP}=\frac{AB}{AP}=\frac{AA_2}{AB}$, so $AA_1=AA_2$, so $A_1=A_2$. Now 
$$\frac{\sin{(b+2c)}}{\sin{(a+2f)}}=\frac{\frac{A_1P}{\sin{(a+2f)}}}{\frac{A_2P}{\sin{(b+2c)}}}=\frac{\frac{CP}{\sin{e}}}{\frac{BP}{\sin{d}}}=\frac{CP\sin{d}}{BP\sin{e}}$$
Similarly, we get $$\frac{\sin{(d+2e)}}{\sin{(c+2b)}}=\frac{AP\sin{f}}{CP\sin{a}}$$ $$\frac{\sin{(f+2a)}}{\sin{(e+2d)}}=\frac{BP\sin{b}}{AP\sin{c}}$$ so multiplying gives the desired equality $$\frac{\sin{(2a+f)}}{\sin{(2f+a)}}\cdot\frac{\sin{(2e+d)}}{\sin{(2d+e)}}\cdot\frac{\sin{(2c+b)}}{\sin{(2b+c)}}=1$$

