Prove that $\gamma = 99^{\circ}$ in this quadrilateral. 
$ABCD$ is a quadrilateral. $\measuredangle{BAD}=86^{\circ}$ and $\measuredangle{CDA}=68^{\circ}$, $|AB|=|CD|$, $E$ and $F$ are midpoints of their segments. $\measuredangle{DEF}=\gamma$. Prove that $\gamma=99^{\circ}$.


My most productive try is defining midpoints of $AB$ and $DC$ as $G$, $H$ respectively and observing the parallelogram $GEFH$. This uses the information of lengths but not the angles. And i couldn't find a way to use angles.
 A: Let $ABKD$ be parallelogram and $G$ be a midpoint of $CK$.
Thus, $FG=\frac{1}{2}BK=b$, $FG||BK$ and $BK||AD$, which says that $EFGD$ is parallelogram.
But $$\measuredangle GDC=\frac{1}{2}\measuredangle CDK=\frac{1}{2}\left(180^{\circ}-86^{\circ}-68^{\circ}\right)=13^{\circ},$$
which says that $$\measuredangle DEF=180^{\circ}-68^{\circ}-13^{\circ}=99^{\circ}$$
and we are done!
A: Draw parallel to $CD$ passing through $A$ and a parallel to $AB$ passing through $D$. Let $BC$ intersect these lines at $G$ and $H$ respectively. Draw the angle bisectors of $\angle GAB$ and $\angle HDC$ and let them intersect $BC$ at $P$ and $Q$ respectively.
Now we'll prove that $AP \parallel EF \parallel DQ$. As $AP \parallel DQ$ and $E$ is the midpoint of $AD$ it's enough to prove that $F$ is the midpoint of $PQ$, i.e. $FP = FQ \iff PB = QC$. Now using the Angle Bisector Theorem  and Sine Theorem we have that:
$$\frac{PB}{QC} = \frac{\frac{AB}{AG} \cdot PG}{\frac{DC}{DH} \cdot QH} = \frac{\frac{PG}{AG}}{\frac{QH}{DH}} = \frac{\frac{\sin \angle PAG}{\sin \angle APG}}{\frac{\sin \angle QDH}{\sin \angle DQH}} = 1$$
Hence the proof.

In the last line we use the fact that: $\angle PAG = \angle QDH = 13^{\circ}$ and $$\sin \angle APG = \sin (\pi - \angle APG) = \sin \angle APB = \sin \angle DQH$$
