This is a question in Waterloo Euclid Contest. I tried to connect $WN$ and extended it to intersect with $x$-axis or constructed $DP$ with $P$ on $x$-axis such that $DP=WA$, but I failed to proceed. The official solution was all about using coordinate geometry or vector method. Is there any method not related coordinate geometry or vector?
I am afraid that my solution is not what you are looking for...
Let $W,A,K,D$ be four distinct complex numbers. Then $M=(W+K)/2$ and $N=(A+D)/2$ and the given equality is equivalent to $$2|N-A|=|(A-W)+(D-K)|=|A-W|+|D-K|$$ Now the Equality of triangle inequality in complex numbers holds iff $(A-W)=s(D-K)$ for some real positive number $s$ which implies that $A-W$ is parallel to $D-K$.