4. The bases $AB$ and $CD$ of the trapezoid $ABCD$ are extended on both sides. The bisectors of the exterior angles of the trapezoid at vertices $A$ and $D$ intersect at point $M$, and the bisectors of the exterior angles at vertices $B$ and $C$ intersect at point $N$. Determine the perimeter of trapezoid $ABCD$ if $MN=2k$.
Solution. Since point $M$ lies on the angle bisectors at vertices $B$ and $C$, it is equidistant from line $A B$ and line $C D$, which means that it lies on the midline of trapezoid $A B C D$. Similarly, point $N$ lies on the midline of trapezoid $A B C D$. Furthermore, triangles $N A P$ and $N D P$ are isosceles (Why?), so $N P=A P=P D$. Similarly, $M Q=B Q=C Q$. Finally, for the perimeter of the trapezoid, we get:

$$
\begin{aligned}
A B+C D+A D+B C & =2 \cdot\left(\frac{A B}{2}+\frac{C D}{2}+A P+B Q\right) \\
& =2 \cdot\left(\frac{A B+C D}{2}+N P+Q M\right) \\
& =2(P Q+N P+Q M) \\
& =2 M N=4 k
\end{aligned}
$$