Let $A_i=(a_i,\frac{1}{a_i})$ and $B_1=(b_1,\frac{1}{b_1})$. Using condition of parallel lines one attains $\frac{b_2-a_2}{\frac{1}{b_2}-\frac{1}{a_2}}=\frac{b_1-a_1}{\frac{1}{b_1}-\frac{1}{a_1}}$ Hence $B_2=(\frac{a_1b_1}{a_2},\frac{a_2}{a_1b_1})$. Similarly $B_3=(\frac{a_1b_1}{a_3},\frac{a_3}{a_1b_1})$, $B_4=(\frac{a_1b_1}{a_4},\frac{a_4}{a_1b_1})$ Using analytic formula for areas of triangles $A_1A_2A_3, A_3A_4A_1$ we can see that it's equal to $\frac{1}{2}\cdot |\left(\frac{a_1}{a_2}+\frac{a_2}{a_3}+\frac{a_3}{a_4}+\frac{a_4}{a_1}-\frac{a_2}{a_1}-\frac{a_3}{a_2}-\frac{a_4}{a_3}-\frac{a_1}{a_4}\right)|$ Generally area of quadrilateral (convex) with vertices $(x_1,y_1),...,(x_4,y_4)$ is equal to $\frac{|x_1y_2+...+x_4y_1-x_2y_1-...-x_1y_4|}{2}$ So you can also count the area of $B_1B_2B_3B_4$ which is exactly the same as for $A_1A_2A_3A_4$