1.67. A line intersects a circle of radius $R$ at points $A$ and $B$ such that $\cup A B=45^{\circ}$, and the line perpendicular to the diameter $A M$ of the circle and passing through its center intersects at point $D$. The line passing through point $B$ and perpendicular to the diameter $A M$ intersects it at point $C$. Find the area of trapezoid $O C B D$.
1.67. By the condition, $\angle B O C=45^{\circ}$ and, consequently, $B C=O C=\frac{R}{\sqrt{2}}$, $A C=R-\frac{R}{\sqrt{2}}=\frac{R(\sqrt{2}-1)}{\sqrt{2}}$ (Fig. 1.53).

Since $\triangle A B C \sim \triangle A D O$, then $\frac{D O}{B C}=\frac{A O}{A C}$,

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-444.jpg?height=586&width=421&top_left_y=907&top_left_x=732)

Fig. 1.53 from which $D O=\frac{A O \cdot B C}{A C}=\frac{R}{\sqrt{2}-1}=R(\sqrt{2}+1)$. We find the area of the trapezoid:

$$
S=\frac{1}{2}(D O+B C) O C=\frac{1}{2}\left(R(\sqrt{2}+1)+\frac{R}{\sqrt{2}}\right) \frac{R}{\sqrt{2}}=\frac{R^{2}(3+\sqrt{2)}}{4}
$$

Answer: $0.25 R^{2}(3+\sqrt{2})$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-445.jpg?height=364&width=578&top_left_y=270&top_left_x=42)

Fig. 1.54