4. The bisectors of the exterior angles at vertices $A$ and $D$ of trapezoid $\mathrm{ABCD}(\mathrm{AB} \| \mathrm{CD}$) intersect at point $\mathrm{M}$, and the bisectors of the exterior angles at vertices B and C intersect at point $\mathrm{N}$ such that $\overline{\mathrm{MN}}=12 \mathrm{~cm}$. Calculate the perimeter of the trapezoid.
Solution: According to the given conditions, points $\mathrm{M}$ and $\mathrm{N}$ are equally distant from the bases. This means that points $\mathrm{M}$ and $\mathrm{N}$ lie on a line that contains the midline of the trapezoid. Then $\measuredangle 1 = \Delta 2$.

- by condition; $\Delta 1 = \triangle 3$ as alternate interior angles. It follows that $\triangle \mathrm{AMP}$ is isosceles, so $\overline{\mathrm{PM}} = \overline{\mathrm{PA}} = \overline{\mathrm{PD}}$. Similarly, $\overline{\mathrm{QN}} = \overline{\mathrm{QB}} = \overline{\mathrm{QC}}$.

$\overline{\mathrm{MN}} = \overline{\mathrm{MP}} + \overline{\mathrm{PQ}} + \overline{\mathrm{QN}} = \frac{\overline{\mathrm{AD}}}{2} + \frac{\overline{\mathrm{AB}} + \overline{\mathrm{CD}}}{2} + \frac{\overline{\mathrm{BC}}}{2}$

$= \frac{1}{2}(\overline{\mathrm{AB}} + \overline{\mathrm{BC}} + \overline{\mathrm{CD}} + \overline{\mathrm{DA}}) = \frac{1}{2} \mathrm{~L}$.

It follows that $\mathrm{L} = 2 \cdot \overline{\mathrm{MN}} = 2 \cdot 12 = 24$; $\mathrm{L} = 24 \mathrm{~cm}$. 

![](https://cdn.mathpix.com/cropped/2024_06_05_e7266f2e82216ac5d245g-3.jpg?height=260&width=663&top_left_y=459&top_left_x=803)