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 :(

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} $