Contruction of a triangle, knowing one angle and it perimeter. What is the locus of a point which splits the perimeter in two at a specific length? Building a triangle (ABC) with a fixed perimeter (P) knowing just one angle (Summit A) is not that easy.
Having fixed one adjacent side (segment AB), we know that the sum of both the others sides is a constant (P-|AB|), and so the third summit (C) is on an ellipse whose focuses are A and B, and passing by the point I called “HalfPerimeter”.

Now, the problem is that I would like to know the locus of a point dividing the perimeter in two equal or not unequal parts.

By tracing it, It appears that if we cut it in two equal parts, the locus of this moving point is the arc of a circle, which we can draw by finding that both the degenerates cases of the triangle are tangent to this circle.

And, if we cut it in two unequal parts, the locus of the moving point is another arc, of a larger circle but with the same center.

However, I have no idea how to prove that this locus is a circle, nor how to prove that all those circles have the same center.
EDIT:
A consequence of this (Corollary of the first part), would be that in the following figure, the perimeter of ABC is independent of the position of M (the moving point), and both ABM and AMC have the same perimeter.
This become obvious, since AB is tangent to the circle in $T_B$, AC is tangent in $T_C$ and BC is tangent in M :
$|AB|+|BM| = |AB|+|BT_B| = (*) = |AC|+|CT_C| = |AC|+|CM|$
(*) those two are independents of M

(But It still doesn't solve the problem with the unequal parts)
 A: 
In this answer, we show how to construct a triangle without much ado using the given data, i.e. its perimeter $p$ and one of its angle $\alpha$, and, then, use it to prove the conjecture OP has stated in the problem statement.
$\underline{\mathrm{1.\space Construction}}$
All we want to know is shown in $\mathrm{Fig.\space 1}$. After drawing the two legs emitting from the vertex $A$ and hemming the given angle $\alpha$, we mark the vertex $C$ on one of the legs a distance $b$ away from $A$. The value of $b$ is arbitrary, but need to satisfy the inequality
$$0 \le b\le \frac{1}{2}p \tag{1}.$$
On the other leg of the angle $\alpha$ we mark point $D$ so that $AD=p=a+b+c$. Then, the point $E$ is taken on $AD$, where $ED=AC=b$. Now, join $EC$ and draw its perpendicular bisector to meet $AD$ at $B$. Since $CBE$ is an isosceles triangle, we have
$$CA+AB+BC=CA+AB+BE=b+a+c=p.$$
Therefore, the perimeter of the $\triangle ABC$ is equal to $P$. We can construct infinite number of such triangles by varying the length $b$ within the limits specified by (1).

$\underline{\mathrm{2.\space A\space Lemma}}$
Before going on to find the locus mentioned in OP’s conjecture, we need to prove the following lemma
$\mathbf{\mathrm{Lemma\space :}}\quad$ All the triangles, which can be constructed using the above-described method, share one $A\mathrm{-excenter}$.
$\mathbf{\mathrm{Proof\space :}}\quad$ We illustrate the proof using $\mathrm{Fig.\space 2}$, where we extend the scenario depicted in $\mathrm{Fig.\space 1}$ by constructing another $\triangle APQ$. The perimeter of $\triangle APQ$ is equal to that of $\triangle ABC$. However, its vertex $Q$ is placed a distance of $d$ further away than the vertex $C$ from the common vertex $A$. As a result, point $T$ is located a distance of $d$ closer than $E$ to the common vertex $A$. The point of intersection of $AD$ and the perpendicular bisector of $QT$ is the vertex $P$.
We have also drawn the angle bisectors $AO$ and $CO$ of $\measuredangle FAD$ and $\measuredangle FCB$ respectively. They intersect each other at $O$, the $A$-excenter of $\triangle ABC$.  To facilitate the proof, it is necessary to draw the auxiliary line segments $OE$, $OQ$, and $OT$ as well.
It is obvious that the perpendicular bisector $MB$ of $CE$ serves also as the angle bisector of $\measuredangle CBE$. Therefore, $MB$ passes through $O$. Our aim is to show that the perpendicular bisector $NP$ of $QT$, which happens to be the angle bisector of $\measuredangle QPD$, also passes through $O$.
Let $\measuredangle BEC$ and $\measuredangle CEO$ equal $\phi$ and $\omega$.
$$\therefore\quad \measuredangle BEO = \phi + \omega. \tag{2}$$
$CBE$ is an isosceles triangle by construction. Therefore, we shall write $\measuredangle ECB = \phi$. $EOC$ is also an isosceles triangles, because $OM$ is the perpendicular bisector of its side $CE$.  So, we have $CO = EO$ and $\measuredangle BCO = \omega$.
We know that $CO$ is the angle bisector of $\measuredangle FCB$. Therefore,
$$ \measuredangle QCO = \measuredangle OCB = \measuredangle ECB + \measuredangle BCO = \phi + \omega. \tag{3}$$
Because of (2), (3), $QC = TE$, and $CO = EO$, the SAS rule can be applied to the pair of triangles $QCO$ and $OTE$ to show that they are congruent, whereby we have $OQ = OT$. This makes $OQT$ an isosceles triangle. Hence, $PN$, which is the perpendicular bisector of $QT$, passes through $O$. Since $PN$ is also the angle bisector of $\measuredangle QPT$, its meeting point $O$ with the angle bisector of $QAP$ is the $A$-excenter of $\triangle APQ$.
This proves that all the triangles, which can be constructed using the above-described method share one $A\mathrm{-excenter}$.

$\underline{\mathrm{3.\space A\space Conjecture}}$
$\mathbf{\mathrm{Conjecture\space :}}\quad$ If a point can be found on the side opposite of the vertex $A$ of certain individual triangles constructed using the above-described method to divide their perimeter in a given ratio, then, those points describe a circular arc having its center at the $A\mathrm{-excenter}$ shared by all those triangles.
$\mathbf{\mathrm{Proof\space :}}\quad$ As shown in $\mathrm{Fig.\space 3}$, an arbitrary point, i.e. $G$, is taken  on the line segment $AD$, which represents the perimeter $p$ of the triangle $ABC$. Let $G$ divide $AD$ so that $AG = \lambda AD$. In order to find the point $H$ that lies on the side $BC$ and divides the perimeter of $\triangle ABC$ such that $AB+BH= \lambda p$, we need to draw a line parallel to $CE$ through $G$ to intersect $BC$.
It is obvious that the perpendicular bisector $BO$ of $CE$ is also the perpendicular bisector of $GH$. Because of this, $OHG$ is an isosceles triangle, so we have
$$OH=OG. \tag{4}$$
The positions of the two points $O$ and $G$ depend only on the size of the perimeter of $\triangle ABC$, which is a constant. Therefore, length of $OG$ is also a constant. Now, according to (4), we can state syllogistically that the length of $OH$, which is equal to $OG$, is also a constant, even if the position of $H$ varies in space from triangle to triangle. Therefore, the locus describes by the perimeter-dividing point $H$ is nothing but a part of a circle with its center at $O$.
$\underline{\mathrm{4.\space Additional\space Information}}$
There is a very simple way to show that radius of the $A$-excircle of all triangles has the same length, if they also have the same perimeter $p$ and the same angle at the vertex $A$. Let this radius be $r_{A}$. Using trigonometry, it can be expressed as
$$r_A = \frac{p}{2}\tan\left(\frac{A}{2}\right).$$
Therefore, if the triangles have the same $p$ and $A$, they have the same $r_A$ as well.
Furthermore, it is possible to express $OH$ as a function of $p$, $A$, and $\lambda$.
$$OH=\frac{p}{2}\sqrt{4\lambda \left(\lambda - 1\right)+\sec^2 \left(\frac{A}{2}\right)}.$$
Since $OH$ does not depend on anything other than the mentioned constants, the locus of $H$ is a circular arc of radius $OH$ with its center at $O$.
You know that the locus coincides with the $A$-excircle when $H$ is the midpoint of $BC$, i.e. $\lambda = 0.5$. If you want to check the validity of (6), you can substitute $\lambda = 0.5$ into it and see whether it reduces to expression (5).
A: So we have a triangle $ABC$, with a fixed angle $\angle BAC$ and fixed perimeter $p=2AH$. Its side $BC$ is then tangent to a circle whose center $O$ lies on the bisector of $\angle BAC$, with $OH\perp AH$.
We take then a point $K\in AH$ and let $B$ vary on $AK$. If we construct $N$ on $BC$ such that $BN=BK$, we also have $ON=OK$, because tangents $BH$ and $BC$, issued from the same point $B$, are symmetric about line $OB$. The locus of $N$ is therefore a circle of center $O$ and radius $OK$.
Note that this is not true if $B$ lies between $K$ and $H$: in that case the locus is a more complicated curve (green loop in figure below).

