Zhendarov R.G.

On the lateral sides $AB$ and $BC$ of an isosceles triangle $ABC$, points $K$ and $L$ are taken respectively, such that $AK + LC = KL$. From the midpoint $M$ of segment $KL$, a line parallel to $BC$ is drawn, and this line intersects side $AC$ at point $N$. Find the measure of angle $KNL$.
Draw a line through point $K$ parallel to $BC$, intersecting the base $AC$ at point $P$. Clearly, $KP = AK$. $MN$ is the midline of the trapezoid (or parallelogram) $KLCP$, so $MN = \frac{1}{2}(LC + KP) = \frac{1}{2}KL$. Therefore, angle $KNL$ subtends the diameter of the circle with center $M$, making it a right angle.

## Answer

$90^{\circ}$.