[

Tangent Circles

[ The perpendicular is shorter than the inclined. Inequalities for right triangles ]

Point $E$ on side $B C$ and point $F$ on side $A D$ of the convex quadrilateral $A B C D$ are positioned such that $B E=2 E C$, $A F=2 F D$. On segment $A E$ is the center of a circle of radius $r$, tangent to sides $A B, B C$ and $C D$. On segment $B F$ is the center of a circle of the same radius $r$, tangent to sides $A B, A D$ and $C D$. Find the area of quadrilateral $A B C D$, given that the specified circles are externally tangent to each other.

#
$A B C D$ - rectangle.

## Solution

Note that $A B \| C D$ (common external tangents to two equal circles). Let $O_{1}$ and $O_{2}$ be the centers of the first and second circles, respectively. Denote $A B=2 x$.

Extend $A E$ and $B F$ to intersect the line $C D$ at points $P$ and $Q$, respectively. From the similarity of triangles $P C E$ and $A B E$, we find that $P C=\frac{1}{2} A B=x$. Similarly, $D Q=x$.

Let the bisector of angle $B C D$ intersect side $A B$ at point $M$. Then triangle $M O_{1} A$ is equal to triangle $C O_{1} P$. Therefore, $A M=C P=x$ and $M$ is the midpoint of $A B$. Hence, $B M=x$.

If $N$ is the point of intersection of the bisector of angle $A D C$ with side $A B$, then similarly we can prove that $N$ is the midpoint of $A B$. Therefore, points $M$ and $T$ coincide.

Since $O_{1} O_{2}$ is the midline of triangle $D M C$, then

$$
C D=2 O_{1} O_{2}=4 r .
$$

$A B C D$ is either an isosceles trapezoid or a rectangle. In the first of these cases, $x>2 r$. Therefore, $A B=2 x > 4 r$. But then $C D<4 r$, which is impossible. Hence, $A B C D$ is a rectangle with sides $2 r, 2 r, 4 r, 4 r$, and its area is $8 r^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_e1837d72990d2467a29dg-26.jpg?height=386&width=1286&top_left_y=1030&top_left_x=386)

## Answer

$8 r^{2}$.