4A. Given a parallelogram $A B C D$. The bisector of $\angle D A B$ intersects side $D C$ at point $L$, and diagonal $B D$ at point $K$, such that $\overline{D K}: \overline{K B}=3: 4$. Calculate the length of segment $L C$, if the perimeter of the parallelogram is 28.
Solution. Since $AL$ is the bisector of $\measuredangle DAB$, we have $\measuredangle DAL = \measuredangle LAB = \measuredangle ALD$, so $\triangle ALD$ is isosceles. Let $\overline{DL} = \overline{AD} = b$ and $\overline{AB} = \overline{DC} = a$. For the areas of the triangles in the diagram, we have

$$
P_{\triangle AKD} : P_{\triangle ABK} = \frac{\overline{AD} \cdot \overline{MK}}{2} : \frac{\overline{AB} \cdot \overline{NK}}{2} = \overline{AD} : \overline{AB} = b : a
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_86f0a3ff472e2cfe9073g-09.jpg?height=235&width=351&top_left_y=707&top_left_x=1132)

where $\overline{MK}, \overline{NK}$ are the corresponding heights dropped to the sides, and for them, $\overline{MK} = \overline{NK}$ because point $K$, as a point on the bisector of $\measuredangle DAB$, is equidistant from the sides of the angle.

On the other hand,

$$
P_{\triangle AKD} : P_{\triangle ABK} = \frac{\overline{DK} \cdot \overline{AH}}{2} : \frac{\overline{BK} \cdot \overline{AH}}{2} = \overline{DK} : \overline{KB} = 3 : 4
$$

Now, equating the ratios of the areas, we get $\frac{b}{a} = \frac{3}{4}$, and from the perimeter of the parallelogram $2(a + b) = 28$, we have $b = 6, a = 8$. Finally, $\overline{LC} = a - b = 2$.