$M$ 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
$$NM + MC = NB. \quad (1)$$
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 $AC$.