[T Concerning Tangent Circles [Trigonometric Ratios in a Right Triangle]

In a convex quadrilateral $A B C D$, two equal circles are inscribed, touching each other. The center of the first circle is on the segment connecting vertex $D$ to the midpoint $E$ of side $A B$, and the center of the second circle is on the segment $C E$. The first circle is tangent to sides $A B, A D$, and $C D$, while the second circle is tangent to sides $A B, B C$, and $C D$. Find the sine of the angle between the diagonals of quadrilateral $A B C D$.
$A B C D-$ is a rectangle.

## Solution

Let $O_{1}$ and $O_{2}$ be the centers of the first and second circles, respectively, $P$ and $Q$ be the points of tangency of the circles with side $C D$, and $r$ be the radius of the circles. Since lines $A B$ and $C D$ are common external tangents to two equal circles, then $A B \| C D$.

Since $D E$ and $C E$ are the angle bisectors of angles $D$ and $C$, triangles $D A E$ and $C B E$ are isosceles. Therefore, $A D = A E = B E = B C$, i.e., $A B C D$ is an isosceles trapezoid or a rectangle. In either case, $D P = C Q$.

Since $P Q = O_{1} O_{2} = 2 r$ and $C D = 2 O_{1} O_{2} = 4 r$ (where $O_{1} O_{2}$ is the midline of triangle $D E C$), then $D P = C Q = r$. Therefore, $D P O_{1}$ and $C Q O_{2}$ are isosceles right triangles. Consequently,

$$
\angle P D E = \angle A D E = \angle Q C E = \angle B C E = 45^{\circ},
$$

Therefore, $A B C D$ is a rectangle. Then $\operatorname{tg} \angle C A B = \frac{B C}{A B} = \frac{1}{2}$ and, if $\varphi$ is the angle between the diagonals of rectangle $A B C D$, then

$$
\varphi = 2 \angle C A B, \sin \varphi = \frac{4}{5}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_f7f9d90261bfc416bd71g-02.jpg?height=438&width=760&top_left_y=364&top_left_x=655)

## Answer

## Problem