[ Area of a triangle (through semiperimeter and radius of inscribed or exscribed circle) [ Similar auxiliary triangles [ Trapezoids (other).

In trapezoid $ABCD$, the bases $AD=4, BC=1$ and the angles $A$ and $D$ at the base are given, equal to $\operatorname{arctg} 2$ and $\operatorname{arctg} 3$, respectively.

Find the radius of the circle inscribed in triangle $CBE$, where $E$ is the point of intersection of the diagonals of the trapezoid.
The radius of the inscribed circle of a triangle is equal to its area divided by the semiperimeter.

## Solution

Drop perpendiculars $B M$ and $C N$ to the base $A D$ of trapezoid $A B C D$. Let $A M=x$. Then $D N=A D-$ $M N-A M=3-x$.

By the condition $B M=A M \operatorname{tg} \angle A=2 x, C N=D N \operatorname{tg} \angle D=3(3-x)$, and since $B M=C N$, we have $2 x=3(3-x)$, from which $x=$ $9 / 5$. Therefore, $A N=A M+M N=14 / 5$,

$D M=11 / 5$.

By the Pythagorean theorem $A C^{2}=A C^{2}=A N^{2}+C N^{2}=(2.8)^{2}+(3.6)^{2}=0.16 \cdot 130=104 / 5, B C^{2}=D M^{2}+B M^{2}=(2.2)^{2}+$ $(3.6)^{2}=89 / 5$.

Let $E K$ be the height of triangle $B E C$. Since triangle $B E C$ is similar to triangle $D E A$ with

the coefficient $B C / A D=1 / 4$, then $E K=1 / 5 B M=18 / 25$,

$C E=\frac{1}{5} A C=\frac{2 \sqrt{130}}{25}, B E=\frac{1}{5} B D=\frac{\sqrt{445}}{25}, 2 S_{B E C}=B C \cdot E K=18 / 25$.

The radius $r$ of the inscribed circle of a triangle is equal to its doubled area divided by the perimeter, hence, $r=\frac{\frac{18}{25}}{1+\frac{2 \sqrt{130}}{25}+\frac{\sqrt{455}}{25}}=\frac{18}{25+2 \sqrt{130}+\sqrt{445}}$.

## Answer

 $\frac{18}{25+2 \sqrt{130}+\sqrt{445}}$.