Let these lines are $y=kx-k+1$ and $y=-\frac{1}{k}x+\frac{1}{k}+1$ with $k < 0$ $A_x,C_x$ are solution of $kx-k+1=\frac{1}{2x}$ or $2kx^2-2(k-1)x -1=0$ so $A_x+C_x=\frac{k-1}{k}, A_xC_x=-\frac{1}{2k}$ $B_x,D_x$ are solution of $ -\frac{1}{k}x+\frac{1}{k}+1=\frac{1}{2x}$ or $\frac{-2x^2}{k}+2(\frac{1}{k}+1)x-1=0$ so $B_x+D_x=k+1, B_xD_x=\frac{k}{2}$ $(C_x-A_x)^2=(A_x+C_x)^2-4A_xC_x= \frac{(k-1)^2}{k^2}+\frac{2}{k}=\frac{k^2+1}{k^2}$ $(D_x-B_x)^2=(D_x+B_x)^2-4B_xD_x=(k+1)^2-2k=k^2+1$ $m=\sqrt{(C_x-A_x)^2(D_x-B_x)^2}=\frac{k^2+1}{k}$ $ABCD$ is non-convex quadrilateral which is triangle $ADC$ minus triangle $ABC$ So $A_{ABCD}=A_{ADC}-A_{ABC}=\frac{FD*AC}{2}-\frac{BD*AC}{2}=\frac{AC*BD}{2}$ $AC^2=(C_x-A_x)^2+(C_y-A_y)^2=(C_x-A_x)^2+(kC_x-kA_x)^2=(k^2+1)(C_x-A_x)^2$ $BD^2=(D_x-B_x)^2( \frac{1}{k^2}+1)$ So $AC^2*BD^2=\frac{(C_x-A_x)^2 (D_x-B_x)^2(k^2+1)^2}{k^2}=m^4$ And so $A_{ABCD}=\frac{m^2}{2}$