In trapezoid $ABCD$ with lateral sides $AB=8$ and $CD=5$, the bisector of angle $B$ intersects the bisectors of angles $A$ and $C$ at points $M$ and $N$ respectively, and the bisector of angle $D$ intersects the same two bisectors at points $L$ and $K$, with point $L$ lying on the base $BC$.

a) In what ratio does the line $MK$ divide the side $AB$, and the line $LN$ - the side $AD$?

b) Find the ratio $KL: MN$, if $LM: KN=4: 7$.
Triangles $A B L$ and $D C L$ are isosceles, $M K\|B C\| A D$, points $M, L, K$ and $N$ lie on the same circle.

## Solution

a) Since $A L$ is the bisector of angle $B A D$, and lines $A D$ and $B C$ are parallel, then $\angle B A L=\angle D A L=\angle B L A$, so triangle $A B L$ is isosceles. Therefore, $B L=A B=8$ and the bisector $B M$ of this triangle is its median and altitude. Similarly, $C L=D=5, L K=K D$ and $C K \perp D L$.

Thus, $M K$ is the midline of triangle $A L D$, so line $M K$ is parallel to the bases of the trapezoid, and point $E$ where line $M K$ intersects line $A B$ is the midpoint of side $A B$.

Let line $L N$ intersect lines $A D$ and $M K$ at points $P$ and $Q$ respectively. Then $A P: P D=M Q: Q K=B L: L C=8: 5$.

b) Segment $L N$ is seen from points $M$ and $K$ at a right angle, so points $M$ and $K$ lie on the circle with diameter $L N$. Therefore,

$\angle L M Q=\angle L M K=\angle L N K=\angle Q N K$ and triangles $L M Q$ and $K N Q$ are similar by two angles. Thus, $L Q: Q K=L M: K N$, from which $L Q=Q K^{L M} /{ }_{K N}=4 / 7 Q K$.

From the similarity of triangles $K Q L$ and $N Q M$, it follows that $\quad K L / M N=L Q / Q M=4 / 7, Q K / Q M=5 / 14$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6409e68af5a1ba5e989dg-58.jpg?height=509&width=738&top_left_y=137&top_left_x=663)

## Answer

a) $1: 1.5: 8 ;$ b) $5: 14$.