## Task 4 - 210734

Given a circle $k$ with center $M$. On $k$ lie the points $A$ and $B$ such that the angle $\angle B M A$ is a right angle. Furthermore, a point $C$ is defined by the following conditions:

(1) $C$ lies on $k$.

(2) It holds that $M B = B C$.

(3) The line through $A$ and $C$ intersects the segment $M B$ at a point $D$.

Determine the size of the angle $\angle C D B$ from these conditions!
Since $B$ and $C$ lie on $k$, it follows that $M B = M C$; by assumption, however, $M B = B C$ as well. Thus, triangle $M C B$ is equilateral, and each of its interior angles is $60^{\circ}$.

If $C$ were on the quarter-circle arc from $A$ to $B$ of $k$, the line through $A$ and $C$ would not intersect the segment $M B$. Therefore, $C$ lies outside this quarter-circle arc, and $\angle A M C = 90^{\circ} + 60^{\circ} = 150^{\circ}$.

Since $A$ and $C$ lie on $k$, triangle $A M C$ is isosceles with $M A = M C$, so $\angle M A C = \angle M C A$. By the angle sum property of a triangle, $\angle M A C = \angle M C A = \frac{1}{2}(180^{\circ} - 150^{\circ}) = 15^{\circ}$.

Thus, by the exterior angle theorem applied to triangle $M C D$,

$$
\angle C D B = \angle C M D + \angle M C D = 60^{\circ} + 15^{\circ} = 75^{\circ}
$$

Alternative solutions: One can also use $\angle C D B = \angle M D A$ (vertical angles), $\angle M D A = 90^{\circ} - 15^{\circ}$ (interior angle in triangle $A D M$).