1.66. First, let's find the side $AB = 2r$ (Fig. 1.52). Let $ED = x$. Then, using the equality $AB + CD = BC + AD$, we get

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-444.jpg?height=359&width=577&top_left_y=105&top_left_x=346)

Fig. 1.52

$2r + CD = \frac{4r}{3} + \frac{4r}{3} + x$, from which $CD = \frac{2r}{3} + x$. In $\triangle CED$, we have $CD^2 = CE^2 + ED^2$, or $\left(\frac{2r}{3} + x\right)^2 = 4r^2 + x^2$, from which $x = \frac{8r}{3}$. Therefore, $CD = \frac{10r}{3}, AD = \frac{12r}{3} = 4r$.

Answer: $2r, \frac{4r}{3}, \frac{10r}{3}$, and $4r$.
1.66. A circle of radius $r$ is inscribed in a rectangular trapezoid.

Find the sides of the trapezoid if its smaller base is equal to $\frac{4 r}{3}$.