## [ Magnitude of the angle between two chords and two secants]

Vertices $B, C, D$ of quadrilateral $A B C D$ are located on a circle with center $O$. This circle intersects side $A D$ at point $E$, and side $A B$ at point $F$. It is known that the chords $B F, F E$, and $E D$ are equal, and the chords $B C$ and $C D$ are also equal to each other. Find the angle $O B C$, given that angle $D A B$ is a right angle.
The picture is symmetrical with respect to the line $A C$. Therefore, $\angle B F E=3 \pi / 4$, and $\angle F B E=\angle F E B=3 \pi / 2-3 \pi / 8=\pi / 8$.

Thus,

$\angle B O F=\angle D O E=\angle E O F=\pi / 4, \angle B O C=\angle C O D=1 / 2(2 \pi-3 \pi / 4)=5 \pi / 8, \quad$ and $\angle O B C=1 / 2(\pi / 2-5 \pi / 16)=3 \pi / 16$.

## Answer

On the side $A B$ of triangle $A B C$, a point $K$ is taken, and on the side $B C$ - points $M$ and $N$ such that $A B=4 A K, C M=B N, M N$ $=2 B N$.

Find the ratios $A O: O N$ and $K O: O M$, where $O$ is the intersection point of the lines $A N$ and $K M$.

## Hint

Prove that $C M: M N: B N=1: 2: 1$. Draw a line through point $N$ parallel to $A C$.

## Solution

Note that point $M$ cannot lie on the segment $B N$ (this would contradict the condition $M N=2 B N$).

Thus, point $N$ lies on the segment $B M$, and by denoting $C M=B N=a$, we find that $M N=2 B N=2 a$. Therefore, $C M: M N: B N=1: 2: 1$.

Draw a line through point $N$ parallel to $A C$. Let $P$ be the point of intersection of this line with side $A B$. By the theorem of proportional segments,

$B P: B K=B N: B M=1: 3, A O: O N=A K: K P=1: 2$.

From the similarity of triangles $B N P$ and $B M K$, we find that $P N=1 / 3 K M$, and from the similarity of triangles $A O K$ and $A N P-O K$ $=\frac{1}{3}$ PN. Therefore, $O K=1 / 9$ KM. Thus, $K O: O M=1: 8$.

## Answer

$1: 2 ; 1: 8$