8.1. Parallelogram $A B C D$ was folded along the diagonal $B D$ so that vertex $C$ remained in place, and vertex $A$ took the position $A^{\prime}$. Segments $B C$ and $A^{\prime} D$ intersected at point $K$, and $B K: K C=3: 2$. Find the area of triangle $A^{\prime} K C$, if the area of parallelogram $A B C D$ is 27.
Solution. As a result of the action described in the condition, $\triangle A B D$ has undergone a symmetry with respect to the line $B D$ and has transformed into $\triangle A^{\prime} B D$. Then $A^{\prime} D = A D = B C$, $A^{\prime} B = A B = C D$. From this, $\triangle A^{\prime} B D = \triangle C D B$ by three sides. But then the heights of these triangles, drawn from the corresponding vertices $A^{\prime}$ and $C$, are equal. Therefore, $A^{\prime} C \| B D$, and $B C$ and $A^{\prime} D$ intersect at point $K$ - not at the midpoint of $B C$, from which $B A^{\prime} C D$ is a trapezoid. Then triangles $A^{\prime} K C$ and $B K D$ are similar with a coefficient of $2 / 3$. In this case, $S_{B K D} = \frac{3}{5} S_{B C D}$, since they have a common height and $B K = \frac{3}{5} B C$. Then $S_{A^{\prime} K C} = \frac{4}{9} S_{B K D} = \frac{4}{9} \cdot \frac{3}{5} S_{B C D} = \frac{4}{9} \cdot \frac{3}{5} \cdot \frac{1}{2} \cdot 27 = 3.6$.

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