Consider trapezoid $ABCD$ with base $AD$ and $BC$. $AB = 9$ and $CD = 5$. Bisector of $A$ intersect bisectors of $B$ and $D$ at points $L$ and $M$. Bisector of $C$ intersect bisectors of $D$ at point $N$ and bisector of $B$ at point $K$, and $K \in AD$ . Find in which proportions $KM$ divide $BC$. And find $\frac{MN}{LK}$ if $\frac{LM}{KN} = \frac{3}{7}$.
Actually it's easy to see that $AB = AK$ and $CD = DK$. So $LN$ lies at middle line. But I stay here for an last hour. Any ideas?
For the first part: triangle $MLN$ is similar to $MAD$, thus line $KM$ divides $LN$ in the ratio $AK:KD$, that is $9:5$. But $KLN$ is similar to $KBC$, so line $KM$ also divides $BC$ in the same ratio $9:5$.
For the second part: the areas of right triangles $KLM$ and $KNM$ are in the ratio $9:5$, because their altitudes, with respect to the common base $KM$, are in that ratio. Hence: $$ {LM\cdot LK\over MN\cdot KN}={9\over5}, \quad\hbox{that is:}\quad {MN\over LK}={5\over9}{LM\over KN}={5\over9}\cdot{3\over7}. $$