Solution:

Let $P$ be the point of intersection between the extension of $CM$ and $BN$ and let $R$ be the point of intersection between the extension of $AM$ and $BC$; let us also denote by $\beta$ the angle $C\widehat{B}A$.

Proof of the first part

We give two proofs of this point.

First argument

By the exterior angle theorem applied to the angle at $N$ of triangle $ABN$, $B\widehat{A}M = B\widehat{A}N = \beta - N\widehat{B}A$. But $C\widehat{B}N + N\widehat{B}A = C\widehat{B}A = \beta$, so $C\widehat{B}N = \beta - N\widehat{B}A = B\widehat{A}M$.

Second argument

Consider the triangles $ABR$ and $BNR$: they have the angle at $R$ in common and the corresponding angles $R\widehat{B}A$ and $B\widehat{N}R$ equal to $\beta$ by hypothesis, hence they are similar. But then also $R\widehat{B}N = R\widehat{A}B$ and hence $C\widehat{B}N = R\widehat{B}N = R\widehat{A}B = B\widehat{A}M$.

Proof of the second part

Consider now the triangle $MNP$: the exterior angle at $M$ is $C\widehat{M}N = C\widehat{M}A$ and is therefore equal to $2\beta$. But by the exterior angle theorem then $M\widehat{P}N = 2\beta - P\widehat{N}M = 2\beta - \beta = \beta$ (hence $MNP$ is isosceles).

Finally consider the triangles $BCP$ and $ABN$: the corresponding angles at $P$ and at $N$ are complements of equal angles by what has just been shown and are therefore equal, while the corresponding angles at $B$ and at $A$ are equal by what was seen in the first part. The two triangles are therefore similar, indeed congruent since the side $BC$ of the first is equal to the corresponding side $AB$ of the second ($ABC$ is isosceles by hypothesis). Hence also $CP = BN$ since they are likewise corresponding sides of these two triangles. But since $MNP$ is isosceles, $CM + MN = CM + MP = CP$ and hence $CM + MN = BN$, as was to be shown.

Denoting by $\alpha$ the angle $A\widehat{B}C$, the locus of points internal to the triangle $ABC$ from which the segment $AC$ is seen under an angle equal to $2\alpha$ is an arc of a circle $\Gamma$ passing through the circumcenter $O$ of $ABC$; indeed the central angle theorem guarantees that $A\widehat{O}C = 2A\widehat{BC}$. Since $ABC$ is isosceles on base $AC$ and acute, $O$ lies along the axis of the segment $AC$ inside the triangle, and the triangles $AOB$, $BOC$, $COA$ turn out to be isosceles. Denoting by $J$ the intersection of $\Gamma$ with $CB$ and setting $\theta = J\widehat{A}M$, $\phi = M\widehat{A}O$, we have:

$\theta = J\widehat{C}M = J\widehat{A}M$ since both subtend the arc $JM$ in $\Gamma$,

$\phi = M\widehat{C}O = M\widehat{A}O$ since both subtend the arc $MO$ in $\Gamma$,

but clearly $\theta + \phi = B\widehat{C}O = \frac{\alpha}{2}$, whence:

$$B\widehat{N}M = \alpha \longrightarrow B\widehat{N}A = C\widehat{O}B = \pi - \alpha,$$

from which follows $A\widehat{B}N = \theta$. Denoting by $\rho$ a reflection with respect to the axis of $AC$ followed by a reflection with respect to the axis of $BA$, let $P = \rho(M)$. Since reflections preserve angles, $PB$ forms an angle equal to $\theta$ with $BA$, which guarantees that $P$ lies on $BN$. The transformation $\rho$, being a composition of axial reflections, is a counterclockwise rotation with center $O$ (intersection of the axis of $CB$ and the axis of $BA$) of amplitude equal to $\pi - \alpha = C\widehat{O}B$; in particular it gives $CM = BP$ and $OM = OP$. Moreover we have:

$$O\widehat{M}A = O\widehat{C}A = O\widehat{A}C, \quad O\widehat{P}N = \pi - O\widehat{P}B = \pi - O\widehat{M}C$$

but $O\widehat{M}A$ and $O\widehat{A}C$ subtend the same arc $CO$ in $\Gamma$ on opposite sides with respect to the center, hence they sum to a straight angle. The triangle $MNP$ is consequently isosceles by congruence of the angles resting on $MP$, and we have:

$$CM + MN = BP + MN = BP + PN = BN$$