Solution:

Since the denominators are nonzero, the equality to be proved is equivalent to $AB \cdot AC = AC \cdot x + AB \cdot x$. Considering then that $x$ is the side of the square, we can rewrite the claim as
$$AB \cdot AC = AC \cdot DE + AB \cdot DF.$$
Now $AB \cdot AC$ is twice the area of $ABC$, $AC \cdot DE$ is twice the area of $ADC$ and $AB \cdot DF$ is twice the area of $ABD$. Since the triangle $ABC$ is the union of the two triangles $ADC$ and $ABD$, which intersect only in the segment $AD$, the claim is proved.

SECOND SOLUTION: The triangle $BDF$ is similar to the triangle $BCA$, since they are both right-angled and share the angle at $B$.
We therefore have the following proportion between the sides
$$BF : BA = DF : CA$$
and since $BF = AB - AF = AB - x$ and $DF = x$, this translates into $(AB - x) : AB = x : AC$. This implies $AB \cdot x = (AB - x) \cdot AC$, that is $AB \cdot x + AC \cdot x = AB \cdot AC$, which is what we wanted to prove.