Locus involving a variable triangle

Question

The base BC of a variable triangle ABC is fixed, and the sum AB+AC is constant. The line DP drawn through the midpoint D of BC parallel to AB meets the line CP at P where CP is a line parallel to the angle bisector of A passing through C. Prove that the locus of P is a circle with D for center.

From the given condition that AB+AC is constant, it is clear that the locus of A must be an ellipse. However constructing an ellipse is not easy and I want to know how to approach the problem without needing to do that. (I don't need a full solution, just a clue)

Answer 1

Consider the figure:

Show that $DE = 0.5AB$ and $EC = 0.5AC$ using the converse of the mid point theorem.
Next, find $\angle EPC$ in terms of $x$ and $\angle B$ (use the parallel lines). Then find $\angle ECP$ and show that the two are equal. This implies $0.5AC = EC = EP$. So, $DP = DE+EP = 0.5AB+0.5AC = 0.5(AB+AC)$ which is a constant.
$D$ is a fixed point, while $P$ is a variable point and $DP$ is constant. The locus of a point which has a fixed distance from a fixed point is a circle.

Answer 2

The above answer is correct. Show that DE=0.5AB and EC=0.5AC using the converse of the mid point theorem. Next, find ∠EPC in terms of x and ∠B (use the parallel lines). Then find ∠ECP and show that the two are equal. This implies 0.5AC=EC=EP . So, DP=DE+EP=0.5AB+0.5AC=0.5(AB+AC) which is a constant. D is a fixed point, while P is a variable point and DP is constant. The locus of a point which has a fixed distance from a fixed point is a circle.