Quadrilaterals with equal sides 

$AC = BD$
$EC = ED$
$AF = FB$
Angle CAF = 70 deg
Angle DBF = 60 deg
We are looking for angle EFA.

I have found through Geogebra that the required angle is 85 deg.
Any ideas how to prove it? I am not so familiar with Geometry :(
 A: Consider the following triangle:-

Let $JM = a, JN = b $ . In this particular $\triangle$, $MK=NL =$ say $x$.
Draw the angle bisector of $\angle J$ , $JO$. 
WLOG $a<b$.
Then , by the internal angle bisector theorem , $MR = k_1a , RN = k_1b , KO= k_2(a+x) , OL = k_2(b+x) $.
Obviously , $MN=k_1(a+b) $ and $KL =k_2(a+b+2x)$
Now , locate a point $J'$ along $JL$ , such that $JJ'=\frac{b-a}{2}$ . While this may seem arbitrary , things will become clear soon.
Draw $J'Q$ parallel to $JO$ .
$J'N=JN-JJ'=\frac{a+b}{2}$.
Using similarity in $\triangle $s $JRN$ and $J'SN$ , $SN$ = $\frac{k_1(a+b)}{2}$
This implies that $S$ is the midpoint of $MN$ !
Similarly , we find $QL$ to equal $k_2(\frac{a+b}{2}+x)$ , proving that $Q$ is the midpoint of $KL$ .
Recall that by construction, $J'Q$ is parallel to $JO$.
Thus , we have discovered the fact , that :-
The line joining the midpoints of the opposite sides of a quadrilateral ,when its other sides are equal , is parallel to the angle bisector of the angle formed by extending the other two sides.
Your problem is now trivial . 
In your case , $\angle MKL=70 , \angle KLN =60 $
$\therefore \angle KJL = 50 \implies \angle RJN = \angle QJ'N = 25$
External angle $J'QK = 25+60 = \boxed{85} $ 
A: Here' is another solution.  Let $\alpha$ and $\beta$ be the angles at $A$ and $B$, resp., let $a$ and $b$ be the length of $AF$ and $AC$, resp.  Consider a coordinate system with origin $F$ and let $FB$ the direction of the abscissa. Let $\varphi$ be the angle in question.
Then 
$$FC=\begin{pmatrix}-a\\0\end{pmatrix}+b\begin{pmatrix}
\cos(\alpha)\\ \sin(\alpha)\end{pmatrix}\text{ and }
FD=\begin{pmatrix}a\\0\end{pmatrix}+b\begin{pmatrix}
-\cos(\beta)\\ \sin(\beta)\end{pmatrix},$$
hence $FE$ the midpoint of $C$ and $D$ is
$$\frac12b\begin{pmatrix}
\cos(\alpha)-\cos(\beta)\\ \sin(\alpha)+\sin(\beta)\end{pmatrix}.$$
Therefore the slope of $FE$ is
$$\tan(180-\varphi)=\frac{\sin(\alpha)+\sin(\beta)}{\cos(\alpha)-\cos(\beta)}
=\frac{2\sin\bigl((\alpha+\beta)/2\bigr)\cos(\bigl((\alpha-\beta)2\bigr)}{-2\sin\bigl((\alpha+\beta)/2\bigr)\sin(\bigl((\alpha-\beta)2\bigr)}
=-\frac{1}{\tan\bigl((\alpha-\beta)/2)\bigr)},$$
that is
$$\tan(\varphi)\cdot\tan\bigl((\alpha-\beta)/2)\bigr)=-1,$$
hence the line $FE$ is perpendicular to one with an angle of $(\alpha-\beta)2$.  Thus, $180-\varphi$ and $(\alpha-\beta)/2$ differ by $90$.
In our case $(\alpha-\beta)/2=5$, so the perpendicular line must have an angle of $95$, that is $180-\varphi=95$.
NB: I'm sure there is a simpler way to achieve this general result.
