smc

Let $AC=x$ and $BF=y$. We have $AFC\sim BFA$ by AA, so $\frac{AF}{FC}=\frac{BF}{AF}$. Substituting in known values gives $\frac{AF}{1}=\frac{y}{AF}$, so $AF=\sqrt{y}$. Also, $AD=x-1$, and using the Pythagorean Theorem on $△ABD$, we have $A{B}^2+(x-1{)}^2=1^2$, so $AB=\sqrt{2x-x^2}$. Using the Pythagorean Theorem on $△AFC$ gives $y+1=x^2$, or $y=x^2-1$. Now, we use the Pythagorean Theorem on $△AFB$ to get $y^2+y=2x-x^2$. Substituting $y=x^2-1$ into this gives $(x^2-1{)}^2+x^2-1=2x-x^2$, or $x^4-2x^2+1+x^2-1=2x-x^2$. Simplifying this and moving all of the terms to one side gives $x^4-2x=0$, and since $x\ne 0$, we can divide by $x$ to get $x^3-2=0$, from which we find that $x=\sqrt[3]{2},\boxed{\sqrt[3]{2}}$.