Let $\square ABCD$ be a convex quadrilateral, and $\overline{\rm CD}=2$. $N$ is the midpoint of $\overline{\rm DB}$. Determine the locus of $N$ when $D$ moves.
What I know is that the locus is the circumference shown(centre F, radius 1), but I don't know why.
On
(Too long for a comment.)
That's an oddly worded question, either carelessly or perhaps deliberately so.
At face value, point $\,A\,$ doesn't figure anywhere, so the quick answer is to take the homothety centered at $\,B\,$ with ratio $\,1/2\,$ which sends $\,D\,$ to $\,N\,$ and $\,C\,$ to the midpoint $\,F\,$ of $\,BC\,$, determine that $\,NF = CD /2 = 1\,$, then conclude that the locus is the circle of radius $\,1\,$ centered at $\,F\,$.
It is certainly true that the locus is a subset of that circle, but on the other hand the problem also states that the quadrilateral $\,ABCD\,$ is convex, and $\,D\,$ "moves". This could be interpreted either as the other three vertices staying put in place, or as the side lengths staying constant with $\,B,C\,$ fixed and $\,A\,$ moving along with $\,D\,$ like in an articulated mechanism. Each interpretation would lead to different answers as to what arcs of the full circle satisfy the condition, neither entirely trivial.
Otherwise put, not enough information for a definitive answer.
Let F’ be the midpoint of BC. We further let $D_1$ and $D_2$ be two different locations of D. Correspondingly, we have $N_1$ and $N_2$ as the midpoints.
Note that $CD_1 = CD_2$. By midpoint theorem, $N_1F’ = (\frac 12)CD_1 = (\frac 12)CD_2 = N_2F’$.
This means F’ is the center of the circle passing through $N_1$ and $N_2$ with radius = half of $CD$.