Problem 8.4. The diagonals of quadrilateral $A B C D$ intersect at point $K$. It turns out that $A B=B K=K D$. On segment $K C$, a point $L$ is marked such that $A K=L C$. Find $\angle B L A$, given that $\angle A B D=52^{\circ}$ and $\angle C D B=74^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ed8f792abd651132e9d3g-3.jpg?height=434&width=509&top_left_y=651&top_left_x=466)
Answer: 42.

Solution. Triangle $ABL$ is equal to triangle $KDC$ ($AL=KC$, $AB=KD$ and $\angle BAK=$ $\angle BKA=\angle DKC$). We have

$$
\begin{aligned}
\angle BLA & =180^{\circ}-\angle BAL-\angle ABL=180^{\circ}-\frac{180^{\circ}-\angle ABD}{2}-\angle CDB= \\
& =90^{\circ}+\frac{1}{2} \angle ABD-\angle CDB=42^{\circ}
\end{aligned}
$$