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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 $c = sqrta^2 + b^2$. 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

$$
beginalign*
fraccolorbluetextbluea - x & = fracba \
fraccolorredtextredx & = fracca \
colorredtextred & = colorbluetextblue
endalign*
$$

Where one easily can solve for $colorbluetextblue$, $colorredtextred$, $x$.

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 $overlineAB$ at point $E$.

Let the measure of $angle ACB$ be $alpha$.
Then $m angle DCE = dfracalpha2$.

Let the line perpendicular to side $overlineAB$ at point $E$ intersect side
$overlineAC$ at point $D$.

Since $overlineED$ is parallel to $overlineBC$, then
$angle ADE cong angle ACB$

By the exterior angle theorem, $mangle DEC = dfracalpha2$.

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 $y = fracx^22b + fracb2$, which you want to intersect with the line $y = fracbax + b$.

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

Let $$CD=DE=y$$ then we get $$fracbc=fracyc-y$$ so $$y=fracbcb+c$$

Not sure if this is less barbaric but using simple trig: $DE=(a-x)tan A$, $DC=fracxcos A$ so the equation to solve is $$(a-x)fracba=fracxsqrta^2+b^2a$$ or $$x=fracabsqrta^2+b^2+b$$

Just another idea to construct point $E$: since $triangleDCE$ is isosceles, it's easy to find $angleACE=(90°-A)/2$