Angle chasing in a square

Question

Attempt:-

I also tried some constructions but couldn't solve it.

Answer 1

Rotate around $A$ for $90^{\circ}$. Then $D$ goes to $B$ and $E$ to new point $E'$.

Then $F,B$ and $E'$ are colinear and triangle $AEF$ is congruent to triangle $AFE'$ (sas) and thus $$x = \angle AE'F = 180 -70-45 = 65$$

Answer 2

Draw diagonal AC and construct $\angle CAH=20$

Now $\triangle ADE$ is congruent to $\triangle BAH$ (A.A.S.)

Then AE = AH since $\angle EAH=40, \angle AHE=AEH=70$.

Now look at quadrilateral AFHE $\angle AFB=70$ and $\angle AEH=70$ then quadrilateral AFHE is cyclic therefore $\angle FAH=\angle FEH=5$

Then you get $x+5=70$

$x=65$