Problem 8.8. In trapezoid $A B C D(A D \| B C) \angle A B C=108^{\circ}$ and $\angle A D C=54^{\circ}$. On ray $B A$ beyond point $A$, point $K$ is marked such that $A K=B C$. Find the angle $D K C$, given that $\angle B K C=27^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ed8f792abd651132e9d3g-5.jpg?height=333&width=398&top_left_y=92&top_left_x=528)
Answer: 18.

Solution. We will prove that triangles $B K C$ and $A D C$ are equal. We have $\angle K B C=\angle D A K$ and $B C=$ $A K$; it remains to show that $B K=A D$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ed8f792abd651132e9d3g-5.jpg?height=498&width=345&top_left_y=627&top_left_x=546)

Fig. 1: to the solution of problem 8.8

For this, we extend $A B$ and $D C$ until they intersect at point $E$ (Fig. 1). Triangle $B E C$ is isosceles, since $\angle B E C=\angle A B C-\angle B C E=\angle A B C-\angle A D C=54^{\circ}=\angle E C B$. Triangle $A E D$ is then also isosceles, as its angles are equal to the angles of triangle $B E C$ due to parallelism. We get $A D=A E=A B+B E=A B+B C=A B+A K=B K$, which completes the justification of the equality of triangles $B K C$ and $A D C$.

Now, knowing that $\angle A K D=\angle B C K$, it is not difficult to calculate the required angle:

$$
\begin{aligned}
\angle D K C & =\angle A K D-\angle B K C=\angle B C K-\angle B K C=\left(180^{\circ}-\angle K B C-\angle B K C\right)-\angle B K C= \\
& =180^{\circ}-\angle A B C-2 \angle B K C=18^{\circ} .
\end{aligned}
$$