32. (5) Given an isosceles triangle \(ABC (AB = BC)\). A point \(K\) is chosen on side \(AB\), and a point \(L\) is chosen on side \(BC\) such that \(AK + CL = \frac{1}{2} AB\). Find the geometric locus of the midpoints of segments \(KL\).
Solution. Mark point $M$ on $AB$ and point $N$ on $BC$ such that $AM = CN = \frac{1}{4} AB$. Also, mark points $M_1$ and $N_1$ as the midpoints of $AB$ and $BC$. On segment $MN$, mark points $M_2$ and $N_2$ as the intersections of segments $AN_1$ and $CM_1$ with segment $MN$. It is not difficult to verify that $M_2M = N_2N = AB / 4$. We will show that the desired locus of points coincides with segment $M_2N_2$. Clearly, $L$ and $K$ lie on opposite sides of line $MN$ and $KM = LN$. Without loss of generality, assume that $K$ lies on segment $BM$. Draw a line through $K$ parallel to $BN$ until it intersects $MN$ at some point $D$. The sides of triangles $MKD$ and $ABC$ are parallel, so $MKD$ is isosceles, $MK = KD$. Segments $KD$ and $NL$ are equal and parallel, hence $KNLD$ is a parallelogram, and the midpoint $E$ of segment $KL$ lies on $DN$; moreover, $NE > N_2N$ and $ME > M_2M$, since if $K'$ is the point symmetric to $L$ with respect to $N$ and $L'$ is the point symmetric to $K$ with respect to $M$, then $NE = K'K / 2 > M_1N_1 / 2 = NN_2$ and $ME = L'L / 2 > M_1N_1 / 2 = MM_2$. Therefore, $E \in M_2N_2$. Conversely, let $E$ be a point on $M_2N_2$, and, say, $EN_2 < EM_2$. By laying off segment $ED = EN$ on $EM$, drawing a line through $D$ parallel to $BN$ until it intersects $BM$ at point $K$, and laying off segment $NL = DK$ on ray $NC$, we obtain the required segment $KL$ with midpoint $E$ (it is not difficult to verify that $L$ will lie on $BC$).