73. On a given circle, two fixed points $A$ and $B$ are taken, $\overrightarrow{A B}=\alpha$. A random circle passes through points $A$ and $B$. A line $l$ is also drawn through $A$, intersecting the circles again at points $C$ and $D$ ($C$ is on the given circle). The tangents to the circles at points $C$ and $D$ (where $C$ and $D$ are the points of tangency) intersect at point $M$; $N$ is a point on $l$ such that $|C N|=|A D|,|D N|=|C A|$. What values can $\angle CMN$ take?
73. We will prove (Fig. 16) that $\triangle C M N$ is similar to $\triangle C A B$. We have $\widehat{M C N}=\widehat{C B A}$. Since the quadrilateral CBDM is inscribed, then

$\frac{|C M|}{|C B|}=\frac{\sin \widehat{C B M}}{\sin \widehat{C M B}}=\frac{\sin \widehat{C D M}}{\sin \widehat{C D B}}=$ $=\frac{\sin \widehat{D B A}}{\sin \widehat{A D B}}=\frac{|A D|}{|A B|}=\frac{|C N|}{|A B|}$.

Thus, $\widehat{C M N}=\widehat{B C A}$, i.e., the required angle is either $\frac{\alpha}{2}$ or

![](https://cdn.mathpix.com/cropped/2024_05_21_6282cf4a54d2a9633c48g-082.jpg?height=350&width=421&top_left_y=460&top_left_x=652)

Fig. 16. $180^{\circ}-\frac{\alpha}{2}$.