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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 $DE\parallel BC$ ($DE$ is parallel to $BC$).

enter image description here

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

enter image description here

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?

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    $\begingroup$ You still have not described what $F$ is either from the statement or from the graph. $\endgroup$ – Hw Chu Apr 17 at 16:42
  • $\begingroup$ Right when I said $F$ i meant $E$. I will fix it in the problem statement =) $\endgroup$ – N3buchadnezzar Apr 17 at 16:47
  • $\begingroup$ cateti is Italian for legs $\endgroup$ – J. W. Tanner Apr 17 at 16:56
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    $\begingroup$ Your system of equations very quickly and easily simplifies to $x=ab/(b+c).$ That does not seem too ugly. But the half-angle method also works. In fact, your problem is a nice way to derive at least one of the half-angle formulas for the tangent function! $\endgroup$ – David K Apr 17 at 20:25
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enter image description here

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

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

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, $m\angle DEC = \dfrac{\alpha}{2}$.

Hence $\triangle EDC$ is isoceles.

So $CD = DE$.

(Added later). Assuming that the sides have lengths of $x$ and $y$, and that $r = \sqrt{x^2+y^2}$, the lengths of the segments are displayed below.

enter image description here

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  • $\begingroup$ This is exactly what I was looking for =) $\endgroup$ – N3buchadnezzar Apr 17 at 18:24
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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 $y = \frac{bc(c - b)}{a^2}$.

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

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  • $\begingroup$ @mathmandan There was an edit after my comment. $\endgroup$ – Michael Biro Apr 17 at 21:17
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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 $$(a-x)\frac{b}{a}=\frac{x\sqrt{a^2+b^2}}{a}$$ 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 $\angle{ACE}=(90°-A)/2$

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