6. (30 points) In triangle $K I A$, angles $K$ and $I$ are equal to $30^{\circ}$. On the line passing through point $K$ perpendicular to side $K I$, a point $N$ is marked such that $A N$ is equal to $K I$. Find the measure of angle $KAN$.
Answer: $90^{\circ}$ or $30^{\circ}$.

Solution: Let $K I=2 a$, and point $H$ be the foot of the perpendicular from vertex $A$ to the

![](https://cdn.mathpix.com/cropped/2024_05_06_3baa4cb0e25778b2cdd7g-3.jpg?height=702&width=391&top_left_y=1622&top_left_x=864)

line $K N$. According to the condition, $\triangle K I A$ is isosceles with base $K I$, hence $A$ lies on the perpendicular bisector of $K I$, from which we have that $A H=a$.

It is easy to notice that $\triangle A H N$ is a right triangle, in which the leg $A H$ is half the hypotenuse $A N$. Therefore, $\angle H A N=60^{\circ}$. Since $A H \| K I$, then $\angle K A H = \angle A K I$. From this, we get that, since point $N$ can be located on different sides of the line $K I$, either $\angle K A N = \angle H A N + \angle H A K = 60^{\circ} + 30^{\circ}$, or $\angle K A N = \angle H A N - \angle H A K = 60^{\circ} - 30^{\circ}$.