Just use trigonometry. By Law of Sines, we can get $\frac{a}{\sin 40^\circ}=\frac{b}{\sin 60^\circ}=\frac{c}{\sin 80^\circ}$. Then, $a(a+b+c)=b(b+c)\Leftrightarrow \sin 40^\circ (\sin 40^\circ +\sin 60^\circ +\sin 80^\circ)=\sin 60^\circ (\sin 60^\circ +\sin 80^\circ)$. We know that $\sin 60^\circ +\sin 80^\circ=2\sin{70^\circ}\cos{10^\circ}=2\sin{70^\circ}\sin{80^\circ}$ and since $40+60+80=180$ we know that $\sin 40^\circ +\sin 60^\circ +\sin 80^\circ=4\cos{20^\circ}\cos{30^\circ}\cos{40^\circ}=4\sin{70^\circ}\sin{60^\circ}\cos{40^\circ}$. Then, we should prove that $4\sin{40^\circ}\sin{70^\circ}\sin{60^\circ}\cos{40^\circ}=2\sin{60^\circ}\sin{70^\circ}\sin{80^\circ}\Leftrightarrow 2\sin{40^\circ}\cos{40^\circ}=\sin{80^\circ}$, which is correct.

Let $D$ be a point on $AC$ such that $CD=a$ and $C\in |AD|$. Let $E$ be a point on $AC$ such that $AE=c$ and $A\in |CE|$. We have $\angle EBD=\angle EBA+\angle ABC+\angle CBD=20^\circ +60^\circ +40^\circ =120^\circ$. $\angle BDA=40^\circ=\angle BAD\Rightarrow BD=BA=c$. Also, $\angle EBC=\angle EBA+\angle ABC=60^\circ+20^\circ=80^\circ=\angle ECB\Rightarrow EB=EC=b+c$. Let $S$ be the area of the triangle $EBD$. $S=\frac{ED\cdot BC\cdot \sin {ECB}}{2}=\frac{(a+b+c)a\sin {80^\circ}}{2}$. $S=\frac{EB\cdot BD\cdot \sin {EBD}}{2}=\frac{(b+c)c\sin {120^\circ}}{2}=\frac{(b+c)c\sin {60^\circ}}{2}=\frac{(b+c)b\sin {80^\circ}}{2}$. Hence, $a(a+b+c)=b(b+c)$.