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I have been working on the following problem

Statement

Assume you have a right angle triangle $Delta ABC$ with cateti $a$, $b$ and hypotenuse $c = sqrt{a^2 + b^2}$. Find or construct a point $D$ on the hypothenuse such that the distance $|CD| = |DE|$, where $E$ is positioned on $AB$ in such a way that $DEparallel BC$ ($DE$ is parallel to $BC$).

Background

My background for wanting such a distance is that I want to create a semicircle from C onto the line $AB$. This can be made clearer in the image below

To be able to make sure the angles is right, I needed the red and blue line to be of same length. This lead to this problem

Solution

Using similar triangles one arrives at the three equations

$$
begin{align*}
frac{color{blue}{text{blue}}}{a - x} & = frac{b}{a} \
frac{color{red}{text{red}}}{x} & = frac{c}{a} \
color{red}{text{red}} & = color{blue}{text{blue}}
end{align*}
$$

Where one easily can solve for $color{blue}{text{blue}}$, $color{red}{text{red}}$, $x$.

Question

I feel my solution is quite barbaric and I feel that there is a better way to solve this problem. Is there another shorter, better, more intuitive solution. Or perhaps there exists a a way to construct the point $D$ in a simpler matter?

Let the angle bisector of $angle ACB$ intersect side $overline{AB}$ at point $E$.

Let the measure of $angle ACB$ be $alpha$.
Then $m angle DCE = dfrac{alpha}{2}$.

Let the line perpendicular to side $overline{AB}$ at point $E$ intersect side
$overline{AC}$ at point $D$.

Since $overline{ED}$ is parallel to $overline{BC}$, then
$angle ADE cong angle ACB$

By the exterior angle theorem, $mangle DEC = dfrac{alpha}{2}$.

Hence $triangle EDC$ is isoceles.

So $CD = DE$.

The curve that traces out the points that are equidistant to $C$ and the line extension of $AB$ is a parabola with focus $C$ and directrix $AB$.

Choosing a convenient coordinate system (parallel to $AB$ with $B$ as the origin), I get the equation $y = frac{x^2}{2b} + frac{b}{2}$, which you want to intersect with the line $y = frac{b}{a}x + b$.

Solving, I get that the point $D$ has $(x,y)$ coordinates of $x = frac{b(b - c)}{a}$ and $y = frac{bc(c - b)}{a^2}$.

Let $$CD=DE=y$$ then we get $$frac{b}{c}=frac{y}{c-y}$$ so $$y=frac{bc}{b+c}$$

Not sure if this is less barbaric but using simple trig: $DE=(a-x)tan A$, $DC=frac{x}{cos A}$ so the equation to solve is $$(a-x)frac{b}{a}=frac{xsqrt{a^2+b^2}}{a}$$ or $$x=frac{ab}{sqrt{a^2+b^2}+b}$$

Just another idea to construct point $E$: since $triangle{DCE}$ is isosceles, it's easy to find $angle{ACE}=(90°-A)/2$