You're correct that there's a better way. However, first note you could more simply, and I believe also easily, explain your goal by stating that with $AB$ being a fixed length, you're trying to determine the position of $B$ on $DE$ so that $AB$ and $BC$ are perpendicular to each other. Thus, this would be equivalent to $C$ and $F$ coinciding. The diagram below shows this, along with a few specified lengths and angles:
In particular, as you stated, $|AB| = a$, $|CE| = b$ and $|DE| = c$ are given. Also, set $|DB| = d$ (so $|BE| = c - d$) and $|AD| = e$ as unknown values.
With $\measuredangle ABD = x$ and $\measuredangle BAD = y$, we have $x + y = 90^{\circ}$. Thus, $\measuredangle CBE = 180^{\circ} - 90^{\circ} - x = 90^{\circ} - x = y$, so $\measuredangle BCE = x$. This means $\triangle ADB \sim \triangle BEC$, as indicated in Sai Mehta's comment.
Next, as suggested by donaastor's comment, using the relations among the similar triangles, plus the Pythagorean theorem, results in a system of several equations, which can be reduced to a polynomial in one of the unknown values. First, the Pythagorean theorem with $\triangle ADB$ gives that
$$d^2 + e^2 = a^2 \tag{1}\label{eq1A}$$
Next, using that $\triangle ADB \sim \triangle BEC$, we get
$$\frac{e}{d} = \frac{c-d}{b} \; \; \to \; \; \; e = \left(\frac{c-d}{b}\right)d \tag{2}\label{eq2A}$$
Substituting this into \eqref{eq1A} results in
$$\begin{equation}\begin{aligned}
d^2 + \left(\left(\frac{c-d}{b}\right)d\right)^2 & = a^2 \\
(b^2)d^2 + (c-d)^2 d^2 & = a^2(b^2) \\
(b^2)d^2 + (c^2)d^2 - (2c)d^3 + d^4 & = a^2(b^2) \\
d^4 - (2c)d^3 + (b^2 + c^2)d^2 - a^2(b^2) & = 0
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Since $a$, $b$ and $c$ are given, this is a quartic equation in the variable $d$. Although \eqref{eq3A} can be solved analytically, as shown in the Wikipedia article, it's usually solved numerically instead, e.g., such as by using the Durand–Kerner method that's also suggested in the article. However, I believe basically all programming languages have either built-in functionality or third-party libraries you can use instead to find the roots (e.g., in Python, there's numpy.roots).
Regardless of how you solve \eqref{eq3A}, with $d$ determined, then $y$ can be calculated in several ways, e.g., by
$$\sin(y) = \frac{d}{a} \; \; \; \to \; \; \; y = \arcsin\left(\frac{d}{a}\right) \tag{4}\label{eq4A}$$
However, especially if you try to solve this with your own code, be careful since quartic equations always have $4$ roots (due to the Fundamental theorem of algebra). I believe there'll usually be $2$ complex conjugates, with the other $2$ being real roots. You want the one with $0 \lt d \lt c$, with the other one being negative, i.e., with $B$ to the left of $D$ (and $A$ below $D$), as shown below:
From \eqref{eq2A}, we then also have that $e \lt 0$, as the diagram above indicates as well.
I assume you always want $B$ to be between $D$ and $E$. If so, especially if you're using your own root-finding method, you should try to ensure you get the correct root (e.g., by choosing an appropriate starting value) or, at the least, add appropriate checking and handling in case you get an out of bound root (i.e., $d \le 0$ or $d \ge c$) instead.
