In parallelogram $A B C D$ with angle $A$ equal to $60^{\circ}$, the bisector of angle $B$ intersects side $C D$ at point $E$. A circle with radius $R$ is inscribed in triangle $E C B$. Another circle is inscribed in trapezoid $A B E D$. Find the distance between the centers of these circles.
The triangle with vertices at point $E$ and the centers of the circles is a right triangle.

## Solution

Let $O_{1}$ and $O_{2}$ be the centers of the circles inscribed in triangle $B C E$ and trapezoid $A B E D$, respectively. Triangle $O_{1} E O_{2}$ is a right triangle because angle $O_{1} E O_{2}$ is a right angle (the angle between the angle bisectors of adjacent angles). Triangle $B C E$ is equilateral

$$
\left(\angle B E C=\angle A B E=\angle E B C=60^{\circ}\right) \text {, }
$$

$O_{1} E=2 R$, the height $E M$ is $3 R$. Therefore,

$$
O_{2} B=E M=3 R \text {. }
$$

Then

$$
E O_{2}=O_{2} B \operatorname{Btg} \angle O_{2} B E=3 R \operatorname{tg} 30^{\circ}=R \sqrt{3}
$$

Therefore,

$$
O_{1} O_{2}=\sqrt{4 R^{2}+3 R^{2}}=R \sqrt{7} \text {. }
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_16ab69fab1743849dba9g-10.jpg?height=389&width=588&top_left_y=2009&top_left_x=735)

Answer

$R \sqrt{7}$.