In trapezoid $ABCD$, the bisectors of angles $A$ and $D$ intersect at point $E$ lying on the side of $BC$. These bisectors divide the trapezoid into three triangles into which the circles are inscribed. One of these circles touches the base $AB$ at the point $K$, and two others touch the bisector $DE$ at points $M$ and $N$. Prove that $BK = MN$.
Let $X=AE\cap DC$ and $Y=DE\cap AB$. Since $DE$ is the angle bisector of $ADC$ we have that $X$ is the reflection of $A$ across $DE$ which implies $|DX|=|AD|$ and similarly $|AD|=|AY|$. Thus $ADXY$ is a rhombus with center $E$. Since $C=BE\cap DX$ with $B$ on $AY$ we have that triangles $BEY$ and $CED$ are congruent which implies $|BE|=|CE|$ and $|DC|=|YB|$. The last inequality implies $|AB|+|CD|=|AB|+|BY|=|AY|=|AD|$. We have $|MN|=||DM|-|DN||=|\frac{|AD|+|DE|-|AE|}{2}-\frac{|DE|+|DC|-|CE|}{2}|=|\frac{(|AD|-|DC|)+|CE|-|AE|}{2}|=\frac{||AB|+|BE|-|AE||}{2}=||BK||=|BK|$.