[The angle between two chords and two secants] [Sum of the angles of a triangle. Theorem about the external angle.]

Vertices $B, C, D$ of quadrilateral $A B C D$ are located on a circle with center $O$, which intersects side

$A B$ at point $F$, and side $A D$ at point $E$. It is known that angle $B A D$ is a right angle, chord $E F$ is equal to chord $F B$, and chords $B C, C D$, $E D$ are equal to each other. Find the angle $A B O$.
Angle $B A D$ is equal to the half-difference of arc $B C D$ and arc $E F$, not containing point $D$.

## Solution

Points $C$ and $D$ divide the arc $B E$, not containing point $F$, into three equal arcs. Let $\beta$ be the measure of each of these three arcs, and $\alpha$ be the measure of each of the two equal arcs $B F$ and $E F$, not containing point $D$. Then $2 \alpha + 3 \beta = 2 \pi$, and since angle $B A D$ is equal to the half-difference of arc $B C D$ and arc $E F$, not containing point $D$, we have

$1 / 2(2 \beta - \alpha) = \pi / 2$.

From the resulting system of equations, we find that $\alpha = \pi / 7$. The desired angle $A B O$ is found from the isosceles triangle BOF:

$\angle A B O = \angle F B O = \pi / 2 - \alpha / 2 = 3 \pi / 7$.

## Answer