Find the ratio in which edges are divided in a tetrahedron

Question

In a regular tetrahedron, E and F are the midpoints of edges AD and BC, respectively.
M and N are on segments CD and EF, respectively, such that α=∠MNC=π/4, β=∠NME=π/3.
In which ratio are EF and CD divided by M and N, respectively?

Taking D as origin: $$ \overrightarrow{DA} = \overrightarrow{a}\\ \overrightarrow{DB} = \overrightarrow{b}\\ \overrightarrow{DC} = \overrightarrow{c}\\ \overrightarrow{DE} = \frac{1}{2}\overrightarrow{a}\\ \overrightarrow{DF} = \frac{1}{2}\overrightarrow{b+c} $$ I don't know how to continue, if you have an easier and more elegant way to do it, don't hesitate showing it, using vector methods.