45. Given a circle with diameter $A B$. A second circle with center at $A$ intersects the first circle at points $C$ and $D$ and the diameter at point $E$. On the arc $C E$, not containing point $D$, a point $M$ is taken, different from points $C$ and $E$. The ray $B M$ intersects the first circle at point $N$. It is known that $|C N|=a,|D N|=b$. Find $|M N|$.
45. Extend $B N$ and $C N$ to intersect the second circle again at points $K$ and $L$ respectively. $|M N|=|N K|$, since $\widehat{A N B}=90^{\circ}$ and $M K$ is a chord of the circle with center at $A$. $\widehat{L N K}=\widehat{B N C}=\widehat{B N D}$ (since the corresponding arcs are equal). Therefore, $|L N|=|N D|=b,|M N| \cdot|N K|=\left|M N i^{2}=a b,\right| M N \mid=$ $=\sqrt{\overline{a b}}$.