Saudi Arabian IMO Booklet

Let $D$ and $E$ be points of rays $ML\to$ and $MK\to$, respectively, such that $LD=AB$ and $KE=AC$. As $\angle DLA=\angle BLM=\angle LBC$, $LD=AB$ and $AL=BC$, triangles $DLA$ and $ABC$ are congruent. Analogously, $EAK$ and $ABC$ are congruent. From $$\angle DAL+\angle LAK+\angle KAE=\angle BCA+\angle CAB+\angle ABC=180^{\circ },$$ it follows that $A$ belongs to the segment $DE$. Therefore, in the light of $\angle LDA=\angle BAC=\angle AEK$, triangle $MDE$ is isosceles with $MD=ME$. Note that $$\begin{array}{rl}BL+LM& =AB+LM-AL=DL+LM-AL=DM-AL,\\ CK+KM& =AC+KM-AK=EK+KM-AK=EM-AK.\end{array}$$ From $DM=EM$ and $AL=AK$ it follows that the right-hand sides of the two equalities above are equal. Hence the left-hand sides are equal as well, i.e. $CK+KM=BL+LM$.