Answer: 30 degrees.

Solution: Let the line on which point $B$ lies be $b$, the line on which point $C$ lies be $c$, and the line on which point $D$ lies be $d$.

Suppose line $A C$ intersects line $b$ at point $M$. By Thales' theorem, $A M = M C$, since the distances between parallel lines are equal. Therefore, $B M$ is the median in the isosceles triangle $A B C$. Thus, $B M$ is also the altitude, meaning that angle $A M B$ is a right angle, as lines $b$ and $c$ are parallel, so $A C$ is perpendicular to line $c$. Let $A C = 2 x$, then the distance between the parallel lines is $x$. Extend $A C$ to intersect line $d$ at point $E$. Then $C E = x$, as the distance between the parallel lines, and $C D = 2 x$, since $A C = 2 x$. Additionally, in triangle $C E D$, angle $E$ is a right angle. Therefore, in the right triangle $C E D$, the hypotenuse is twice the length of one of the legs. This means that angle $E C D$ is 60 degrees, and the adjacent angle $A C D$ is 120 degrees. Then, in the isosceles triangle $A C D$, the two acute angles are each 30 degrees.

Criteria: It is proven that $A C$ is perpendicular to line $c-2$ points.

![](https://cdn.mathpix.com/cropped/2024_05_06_ee373ffab09e7a30d5bdg-2.jpg?height=402&width=551&top_left_y=790&top_left_x=1386)
8.3. Find the angle $D A C$, given that $A B=B C$ and $A C=C D$, and the lines on which points $A, B, C, D$ lie are parallel, with the distances between adjacent lines being equal. Point $A$ is to the left of $B$, $C$ is to the left of $B$, and $D$ is to the right of $C$ (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_ee373ffab09e7a30d5bdg-2.jpg?height=325&width=488&top_left_y=397&top_left_x=687)