Find the length x such that the two distances in the triangle are the same

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 $$BC$$). 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?

• You still have not described what $F$ is either from the statement or from the graph. – Hw Chu Apr 17 at 16:42
• Right when I said $F$ i meant $E$. I will fix it in the problem statement =) – N3buchadnezzar Apr 17 at 16:47
• cateti is Italian for legs – J. W. Tanner Apr 17 at 16:56
• 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! – David K Apr 17 at 20:25 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. • This is exactly what I was looking for =) – N3buchadnezzar Apr 17 at 18:24

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 $$y=\frac{bc}{b+c}$$

• @mathmandan There was an edit after my comment. – Michael Biro Apr 17 at 21:17

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 $$\angle{ACE}=(90°-A)/2$$