On the segment $A B$, a point $C$ is taken, and segments $A B$ and $C B$ serve as diameters of circles. Chord $A M$ is tangent to the smaller circle at point $D$. Line $B D$ intersects the larger circle at point $N, \angle D A B=\boldsymbol{\alpha}, A B=2 R$. Find the area of the quadrilateral $A B M N$.
Let $O$ be the center of the circle with diameter $B C$. Then $O D \| B M$. Next, use the formula for the area of a quadrilateral through its diagonals and the angle between them.

## Solution

Let $O$ be the center of the circle with diameter $B C$. Since $B M \perp A M$ (point $M$ lies on the circle with diameter $A B$) and $O D \perp A M$ (the radius to the point of tangency is perpendicular to the tangent), then $B M \| O D$.

From this, it follows that $\angle O B D = \angle O D B = \angle M B D$, i.e., $B D$ is the bisector of angle $A B M$, which is $90^{\circ} - \alpha$. Therefore,

$$
\angle A B N = \frac{1}{2} \angle A B M = 45^{\circ} - \frac{\alpha}{2}
$$

Thus,

$$
B N = A B \cdot \cos \angle A B N = 2 R \cos \left(45^{\circ} - \frac{\alpha}{2}\right)
$$

From the right triangle $A M B$, we find that

$$
A M = A B \cdot \cos \angle B A M = 2 R \cos \alpha
$$

Since $A D N$ is the exterior angle of triangle $A D B$, then

$$
\angle A D N = \angle B A D + \angle A B N = \alpha + 45^{\circ} - \frac{\alpha}{2} = 45^{\circ} + \frac{\alpha}{2}
$$

Therefore,

$$
\begin{gathered}
S_{\mathrm{ABMN}} = \frac{1}{2} \cdot A M \cdot B N \cdot \sin \angle A D N = \frac{1}{2} \cdot 2 R \cos \alpha \cdot 2 R \cos \left(45^{\circ} - \frac{\alpha}{2}\right) \cdot \sin \left(45^{\circ} + \frac{\alpha}{2}\right) = \\
= 2 R^{2} \cdot \frac{1}{2} \left(\sin 90^{\circ} + \sin \alpha\right) = R^{2} \cos \alpha (1 + \sin \alpha)
\end{gathered}
$$

## Answer

$R^{2} \cos \alpha (1 + \sin \alpha)$