In triangle $\triangle ABC$, $D$ and $E$ lie on sides $CA$ and $AB$ such that $BE = 6$ and $CD = 10$. Let $M$ and $N$ be the midpoints of segments $BD$ and $CE$, respectively. If $MN = 7$, then what is the measure of $\angle BAC$?
I'm not sure how to start this, I can't find any similar or parallel lines in the figure. I can construct the trapezoids but I do not know how to utilize them.
If $A=E=D$, then triangle $MAN$ has sides $3$, $5$ and $7$. From the cosine rule it follows that $$ \cos\angle MAN={3^2+5^5-7^2\over 2\cdot 3\cdot 5}=-{1\over2}, \quad\text{that is:}\quad \angle MAN=\angle BAC=120°. $$ This is a particular case, but the problem implies that the same solution must hold also in the general case.