Find $AD\cdot AF$ when point $C$ is reflected across $AD$

Question

$ABC$ is a triangle with side lengths $AB=AC=20$ and $BC=18$. $D$ is a point on $BC$ such that $BD<CD$. $C$ is reflected across $AD$ and lands on $E$. $EB$ and $AD$ meet at $F$. Find $AD\cdot AF$. No progress at all. This is my first question where I have seen points getting reflected and I really have no idea how to approach this question. Any help would be appreciated. Also is there any sort of "name" given to these categories of problems?

Edit: The question is indeed correct and the answer is $400$. But I have no clue how to get that value.

Here is a Geogebra diagram: (not completely accurate) https://www.geogebra.org/classic/hecrtmsd

Answer 1

Note that $AE=AC=AB\implies \angle ACF= \angle AEB=\angle ABE\implies F\in\odot(ABC)$. Now, note that $$\angle ABC=\angle ACB=\angle AFB\implies AB \text{ is tangent to } \odot(BDF) \text{ at $B$}$$Thus, by power of point, $$400=AB^2=\mathcal P_{\odot(BDF)}(A)=AD\times AF$$