I need some help on this problem on Geometric Loci. The base $BC$ of a variable triangle is fixed, and the sum $AB + BC$ is constant, the line drawn through the midpoint $D$ of $BC$ parallel to $AB$ meets the parallel $CP$ through $C$ to the internal bisector of the angle at $A$, in $P$. Prove that the locus of $P$ is a circle having $D$ for centre.
Here are my assumptions, since the base $BC$ if fixed, it follows that $AB$ must lie on a circle with centre $B$. But I don't know if this is gonna help as that would leave me with three possibilities for which $AB = BC$ or $AB < BC$ or $AB > BC$ and considering any of these cases seems to lead me to a dead end. 
 A: My answer will assume $AB+AC$ is constant as opposed to the original statement (since by Jack's answer, under the original statement the locus of $P$ is not on a circle). Under this condition we can prove that the locus of $P$ is indeed a circle.
Let us find the length of $PD$ and try to establish it as a constant function of $BC$ and $AB+AC$. Let $E$ be the intersection of the angle bisector and line $PD$, and let $F$ be the intersection of the angle bisector and line $BC$.  Notice that $$\frac{PD}{CD} = \frac{DE}{DF} = \frac{AB}{BF}\implies PD = \frac{BC}{2}\cdot \frac{AB}{BF}$$
Because we also have$$\frac{AB}{AC}=\frac{BF}{FC}\implies \frac{AB}{AB+AC}=\frac{BF}{BF+FC}\implies \frac{AB}{BF} = \frac{AB+AC}{BC}$$
We know that $$PD = \frac{AB+AC}{2}$$which is constant. This implies that the locus of $P$ is a circle with center $D$ and radius $\frac{AB+AC}{2}$.
A: Something's odd since it looks like $P$ moves on a cardioid, not on a circle:

Simple trigonometry prove this in a straightforward way: it is enough to take $D$ as the origin, assume $A=-1+re^{i\theta}$, compute the slope of the bisector from $A$ and intersect a couple of lines.
Things get interesting if $\color{red}{AB+AC}$ is constant, instead. In this case $A$ lies on a ellipse $\mathcal{E}$ with foci at $B,C$ and center at $D$. By the ellipse optical property the angle bisector from $A$ is orthogonal to the tangent $\tau$ to $\mathcal{E}$ at $A$:

Let $\tau$ be the perpendicular through $A$ to the internal angle bisector and $B'$ be the symmetric of $B$ with respect to $\tau$. For any $A'\in\tau$ such that $A'\neq A$ we have
$$ A'B+A'C= B'A'+A'C > B'C = B'A+A C = AB+AC $$
hence $\tau\cap\mathcal{E}=\{A\}$. By angle chasing $P\in\tau$ and by Thales' theorem $PD$ goes through the midpoint of $N$ of $AC$. In particular:
$$PD =PN+ND = \frac{1}{2}AC+\frac{1}{2}AB $$
is constant and $P$ lies on a circle centered at $D$.
