Let $K$ and $P$ be tangent points of an inscribed circle of $ABCD$ to $AB$ and $BC$ respectively (Fig. 38). It is that $\triangle C_1PO = \triangle A_1KO$, since they are right triangles with equal legs $KO = PO$ and hypotenuses $C_1O = A_1O$ as radii of respectful circles. Thus we have $C_1P = A_1K$. Since $BK = BP$, as two tangent segments that are drawn from the same external point, then $C_1B = A_1B$. Similarly, $C_2D = A_2D$. Moreover, from the discussion above we find that $AB = BC$ and $AD = CD$. Therefore, $ABCD$ is a kite, whose diagonal $BD$ is a line of symmetry, hence $B$, $D$ and $O$ lie on the same line.

Line $BD$ passes through the points $B$ and $O$. Those points are equidistant from the endpoints of segment $A_1C_1$, hence $B$ and $O$ belong to its perpendicular bisector. Similarly, line $BD$ is a perpendicular bisector of $A_2C_2$. Since $BD \perp A_1C_1$ and $BD \perp A_2C_2$, then $A_1C_1 \parallel A_2C_2$.

Next, we consider two cases.

If $A_1A_2 \parallel C_1C_2$, then the quadrilateral $A_1A_2C_2C_1$ is a parallelogram, then $A_1A_2C_2C_1$ is a rectangle. All three lines $A_1A_2$, $C_1C_2$ and $BD$ are perpendicular to $A_1C_1$, hence they are parallel.

If $A_1A_2$ and $C_1C_2$ are not parallel, then the quadrilateral $A_1A_2C_2C_1$ is a trapezoid. Since a line $BD$ passes through midpoints of trapezoid bases, then it contains a point of intersection of extensions of legs $A_1A_2$ and $C_1C_2$. Hence lines $A_1A_2$, $C_1C_2$ and $BD$ intersect at one point.