Let us invert in $C_1$ with any radius $R$, so that $(B_1C_1D_1)$ becomes a line parallel to $A_2A_3$ and $A_3^*A_2^*B_1^*D_1^*$ becomes an isosceles trapezoid. By inverted distance formula, we have $$\frac{A_2D_1}{A_3D_1}=\frac{A_2^*D_1^*}{A_3^*D_1^*}\frac{R^2/(C_1A_2^*\cdot C_1D_1^*)}{R^2/(C_1A_3^*\cdot C_1D_1^*)}=\frac{B_2^*A_3^*}{B_2^*A_3^*}\frac{C_1A_3^*}{C_1A_2^*}=\frac{B_1A_3}{B_1A_2}\frac{R^2/(C_1B_1\cdot C_1A_3)}{R^2/(C_1B_1\cdot C_1A_2)}\frac{R^2/C_1A_3}{R^2/C_1A_2}=\frac{B_1A_3}{B_1A_2}\left(\frac{C_1A_3}{C_1A_2}\right)^2$$By ceva and trig ceva, $$\prod_{cyc}\frac{\sin(A_2A_1D_1)}{\sin(DA_1A_3)}=\prod\frac{A_2D_1}{A_3D_1}=\prod \frac{B_1A_3}{B_1A_2}\left(\prod \frac{C_1A_3}{C_1A_2}\right)^2=1\cdot 1^2=1$$Therefore, by trig ceva, $A_1D_1,A_2D_2,A_3D_3$ concur.