66th Belarusian Mathematical Olympiad

$M$ is the intersection point of the line $\ell \parallel AC$ passing through the intersection point of the bisector of the angle $ACB$ and the side $AB$.

Let $M$ be the point we search for and $NM\parallel AC$ (see the Fig.).

Let $P(CMNA)$ denote the perimeter of the trapezoid $CMNA$. Then $P(CMNA)=AN+NM+MC+CA$ and, by condition, $P(CMNA)=AB+AC=AN+NB+AC$, whence $$NM+MC=NB.(1)$$ Since $MN\parallel AC$, the triangles $NBM$ and $ABC$ are similar, so $NB:NM=AB:AC=2$, i.e., $NB=2NM$. From (1) it follows that $MC=MN$, hence, the triangle $NMC$ is isosceles. Therefore, $\angle MNC=\angle MCN$. Since $NM\parallel AC$, we have $\angle MNC=\angle NCA$. Thus, $\angle MCN=\angle NCA$, i.e., $NC$ is the bisector of the angle $ACB$ of the given triangle $ABC$.



The construction of the required point $M$: draw the bisector of the angle $ACB$ (the standard rule-compass construction) and let $N$ be the intersection point of this bisector and the side $AB$. Draw the line $\ell$ passing through $N$ parallel to $AC$ (the standard rule-compass construction). The required point $M$ is the intersection point of the line $\ell$ and the side $BC$. Indeed, it is easy to see that the perimeter of the trapezoid $ANMC$ thus obtained is equal to the sum of the lengths of the sides $AB$ and $AC$.