[ Rectangles and squares. Properties and characteristics ]

Side $A B$ of rectangle $A B C D$ is equal to 12, and side $A D$ is equal to 5. The diagonals of the rectangle intersect at point $E$. Find the ratio of the distance from point $E$ to the center of the circle inscribed in triangle $A E D$ to the distance from point $E$ to the center of the circle inscribed in triangle $D E C$.

#
The distance from the vertex of a triangle to the nearest point of tangency with the inscribed circle is equal to the difference between the semiperimeter and the opposite side.

## Solution

From the properties of a rectangle, it follows that $D E=\frac{1}{2} D B=\frac{13}{2}$. Let $O_{1}$ and $O_{2}$ be the centers of the given circles, $M$ and $N$ — the points of tangency of the circles with the segment $D E, P$ and $Q$ - the midpoints of $A D$ and $D C$. Denote $\angle P E D=\alpha$. Then

$$
\angle N O_{2} E=\alpha, E O_{1}=\frac{M E}{\cos \alpha}, E O_{2}=\frac{N E}{\sin \alpha} .
$$

Therefore,

$$
\begin{gathered}
\frac{E O_{1}}{E O_{2}}=\frac{M E}{N E} \operatorname{tg} \alpha, M E=\frac{13}{2}+\frac{5}{2}-5=4 \\
N E=\frac{13}{2}+6-12=\frac{1}{2}, \operatorname{tg} \alpha=\frac{5}{12}
\end{gathered}
$$

Thus, $\frac{E O_{1}}{E O_{2}}=\frac{10}{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_16ab69fab1743849dba9g-13.jpg?height=312&width=585&top_left_y=2122&top_left_x=742)

## Answer

$\frac{10}{3}$

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