From $A$ to $E$

We can consider the vertices of the nonagon as the images of the 9th roots of unity on the Argand plane, as in the figure.
Let $A$ , $B$ , $C$ , $D$ , and $E$ be the images of ${{\zeta }_{k}}=\cos \frac{2k\pi }{9}+i\sin \frac{2k\pi }{9}$ , for $k=0,1,2,3,4$ respectively. Then,

$\begin{aligned} {{a}^{2}} & ={{\left| \overline{AB} \right|}^{2}} \\ & ={{\left| {{\zeta }_{1}}-{{\zeta }_{0}} \right|}^{2}} \\ & =\left( {{\zeta }_{1}}-{{\zeta }_{0}} \right)\left( \overline{{{\zeta }_{1}}-{{\zeta }_{0}}} \right) \\ & =\left( {{\zeta }_{1}}-1 \right)\left( \overline{{{\zeta }_{1}}}-1 \right) \\ & ={{\zeta }_{1}}\overline{{{\zeta }_{1}}}-\left( {{\zeta }_{1}}+\overline{{{\zeta }_{1}}} \right)+1 \\ & ={{\left| {{\zeta }_{1}} \right|}^{2}}-\left( {{\zeta }_{1}}+\overline{{{\zeta }_{1}}} \right)+1 \\ & =2-\left( {{\zeta }_{1}}+\overline{{{\zeta }_{1}}} \right) \\ & =2-2\cos \frac{2\pi }{9} \\ \end{aligned}$ Similarly, ${{b}^{2}}={{\left| \overline{AC} \right|}^{2}}=2-2\cos \frac{4\pi }{9} \ \ \ \ \ {{c}^{2}}={{\left| \overline{AD} \right|}^{2}}=2-2\cos \frac{6\pi }{9} \ \ \ \ \ {{d}^{2}}={{\left| \overline{AE} \right|}^{2}}=2-2\cos \frac{8\pi }{9}$ Using these expressions we have $\begin{aligned} {{b}^{2}}-{{a}^{2}} & =\left( 2-2\cos \frac{4\pi }{9} \right)-\left( 2-2\cos \frac{2\pi }{9} \right) \\ & =2\left( \cos \frac{2\pi }{9}-\cos \frac{4\pi }{9} \right) \\ & =2\times 2\sin \frac{\frac{2\pi }{9}+\frac{4\pi }{9}}{2}\sin \frac{\frac{4\pi }{9}-\frac{2\pi }{9}}{2} \\ & =4\sin \frac{3\pi }{9}\sin \frac{\pi }{9} \ \ \ \ \ (1)\\ \end{aligned}$ Similarly, ${{c}^{2}}-{{a}^{2}}=4\sin \frac{4\pi }{9}\sin \frac{2\pi }{9} \ \ \ \ \ (2)$ Combining $(1)$ and $(2)$ , we find $\alpha =\left( {{b}^{2}}-{{a}^{2}} \right)\left( {{c}^{2}}-{{a}^{2}} \right)=16\sin \frac{\pi }{9}\sin \frac{2\pi }{9}\sin \frac{3\pi }{9}\sin \frac{4\pi }{9} \ \ \ \ \ (3)$ Similarly, we get $\beta =\left( {{d}^{2}}-{{a}^{2}} \right)\left( {{d}^{2}}-{{a}^{2}} \right)=16\sin \frac{7\pi }{9}\sin \left( -\frac{\pi }{9} \right)\sin \frac{5\pi }{9}\sin \left( -\frac{2\pi }{3} \right) \ \ \ \ \ (4)$ But, $\sin \frac{7\pi }{9}=\sin \frac{2\pi }{9}, \ \ \ \ \ \ \ \sin \left( -\frac{\pi }{9} \right)=-\sin \frac{\pi }{9}, \ \ \ \ \ \ \ \sin \frac{5\pi }{9}=\sin \frac{4\pi }{9}, \ \ \ \ \ \ \ \ \sin \left( -\frac{2\pi }{3} \right)=-\sin \frac{3\pi }{9}$ Hence, $(3)$ and $(4)$ give $\boxed{\alpha=\beta}$ .