Problem 1005

Points $C_{1}$ and $D_{1}$ lie on the perpendicular bisector of segment $A B$, so $A B \perp C_{1} D_{1}$. Similarly, $C_{1} D_{1} \perp$ $A_{2} B_{2}$, which means $A B \parallel A_{2} B_{2}$. Similarly, it can be proven that the corresponding sides and diagonals of quadrilaterals $A B C D$ and $A_{2} B_{2} C_{2} D_{2}$ are parallel. Therefore, these quadrilaterals are similar.

Let $M$ be the midpoint of segment $A C$. Then $B_{1} M=|A M \operatorname{ctg} D|$ and $D_{1} M=|A M \operatorname{ctg} B|$, and $B_{1} D_{1}=|\operatorname{ctg} B+\operatorname{ctg} D| \cdot A C / 2$. If we rotate quadrilateral $A_{1} B_{1} C_{1} D_{1}$ by $90^{\circ}$, then, using the result of problem 6.25, we get that this quadrilateral is convex, and $\operatorname{ctg} A=-\operatorname{ctg} C_{1}$, etc. Therefore, $A_{2} C_{2}=|\operatorname{ctg} A+\operatorname{ctg} C| \cdot B_{1} D_{1} / 2=|(\operatorname{ctg} A+\operatorname{ctg} C)(\operatorname{ctg} B+\operatorname{ctg} D) / 4| \cdot A C$.