Question

10
nq = mq
ne/qf =?
11

Explanation:

Step1: Consider similar - triangles

Since $\angle NFM=\angle NEM = 90^{\circ}$, points $N$, $F$, $E$, $M$ are concyclic.

Step2: Use angle - relationships

In right - triangle $NMQ$, because $NQ = MQ$, $\angle QNM=\angle QMN = 45^{\circ}$.
In right - triangle $NEM$, $\angle NME = 75^{\circ}$, so $\angle ENM=15^{\circ}$. Then $\angle ENQ=\angle QNM-\angle ENM = 45^{\circ}-15^{\circ}=30^{\circ}$.

Step3: Prove similarity of triangles

$\triangle NEF\sim\triangle QMF$ (by AA similarity, as $\angle NFE=\angle QFM = 90^{\circ}$ and $\angle ENF=\angle MQF$ because angles in the same segment of the circum - circle of $NFEM$).
Let's consider right - triangle $NQF$. If we assume $\angle NQF = 30^{\circ}$ (from the above angle calculations). In right - triangle $NEF$ and right - triangle $QMF$, we know that $\frac{NE}{QF}=\frac{NF}{MF}\times\frac{EF}{MF}$.
Since $\angle ENQ = 30^{\circ}$ in right - triangle $NEQ$ and $\angle NQF$ is also related to the angles in the figure, and considering the properties of right - triangles and similar triangles:
In right - triangle $NEQ$, if we consider the trigonometric relations. Let's assume $NQ = 2x$. In right - triangle $NQF$, if $\angle NQF = 30^{\circ}$, and in right - triangle $NEQ$ with $\angle ENQ = 30^{\circ}$.
We know that in right - triangle $NEQ$, $NE=\frac{1}{2}NQ$ and in right - triangle $NQF$, $QF=\frac{\sqrt{3}}{2}NQ$.
So $\frac{NE}{QF}=\frac{\frac{1}{2}NQ}{\frac{\sqrt{3}}{2}NQ}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$.

Answer:

$\frac{\sqrt{3}}{3}$