Csdrhrt

$AC = BD$

$EC = ED$

$AF = FB$

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 :(

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} $

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
$$FC=begin{pmatrix}-a\0end{pmatrix}+bbegin{pmatrix}
cos(alpha)\ sin(alpha)end{pmatrix}text{ and }
FD=begin{pmatrix}a\0end{pmatrix}+bbegin{pmatrix}
-cos(beta)\ sin(beta)end{pmatrix},$$
hence $FE$ the midpoint of $C$ and $D$ is
$$frac12bbegin{pmatrix}
cos(alpha)-cos(beta)\ sin(alpha)+sin(beta)end{pmatrix}.$$
Therefore the slope of $FE$ is
$$tan(180-varphi)=frac{sin(alpha)+sin(beta)}{cos(alpha)-cos(beta)}
=frac{2sinbigl((alpha+beta)/2bigr)cos(bigl((alpha-beta)2bigr)}{-2sinbigl((alpha+beta)/2bigr)sin(bigl((alpha-beta)2bigr)}
=-frac{1}{tanbigl((alpha-beta)/2)bigr)},$$
that is
$$tan(varphi)cdottanbigl((alpha-beta)/2)bigr)=-1,$$
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.