$ABC$ and $BCD$ are triangles, $AB\perp AC$, $\measuredangle{ACB}=\measuredangle{ACD}$, $2|BE|=5|DF|$, $|BC|=20$, $|CD|=5$, $|AE|=x$ is given. Find $x$.
Here is a diagram for it:
First, i assigned some varibles via given relationship $2|BE|=5|DF|$:
$$
|BE|=5k, |DF|=2k
$$
According to angle bisector theorem i can write:
$$
\frac{5}{2k}=\frac{20}{5k+EF}\Rightarrow |EF|=3k
$$
I think it is appropriate to draw a segment that is parallel to $AC$, so it will divide $BC$ into two segments proportional to $5k,3k$. After these, i couldn't continue. Seems like $\measuredangle{BCA}=\measuredangle{EAF}$ but, i counldn't prove it. Thanks for your effort.
Let $M$ be the midpoint of $BC$. My goal is to prove that $AEM$ is on the same line, and use this to find the value of $x$. Notice that by the angle bisector theorem, $$\frac{DF}{BF} = \frac{CD}{CB} = \frac{1}{4}\implies BD = 5DF = 2BE\implies \text{$E$ is the midpoint of $BD$}$$ Further, let $AM$ meet $BD$ at $E'$, then $$\angle MAC = \angle ACM = \angle ACD\implies AM\parallel CD\implies \frac{BE'}{BD} = \frac{BM}{BC}=\frac{1}{2}\implies \text{$E'$ is midpoint of $BD$}$$ Hence, $E=E'$. We can then get $x$ by $$x=AE=AM-ME = \frac{BC}{2}-\frac{CD}{2}=\frac{15}{2}$$