Length of line perpendicular to $\overline{AB}$ ($A = (0, y_1)$, $B = (x_1, 0)$) intersecting $AB$ at $1/5$ its length, and y axis at $(0, y_2)$

Question

I'm feeling pretty stupid that I've spent some time on this and am still unsure of the answer, but it's been quite a while since I've done any trigonometry so if someone wouldn't mind explaining how to find a solution it would be very much appreciated!

The right-angled triangle, $\triangle ABC$, has coordinates, $A = (0,y_1)$, $B = (x_1,0)$, and $C = (0,0)$, where $x_1$ and $y_1$ are unknown positive values. The hypotenuse has length, $\overline{AB}$ = $5a$ (where the value of $a$ is arbitrary).

The line, $\overline{DE}$, runs perpendicular to $\overline{AB}$, with the point, $D$, lying $1/5$ of the way along $\overline{AB}$, and $E$ = $(0,0.141)$.

What are the two possible values of the length, $\overline{DE}$, and angle, $\angle ABC$?

Answer 1

Recognize that the right triangles ABC and AED are similar, we have

$$\frac {AE}{AD} = \frac{AB}{AC}$$

Substitute $AE = y_1-y_2$, $AC = y_1$ and $AD = \frac15AB = a$ into above ratio,

$$\frac{y_1-y_2}{a} = \frac{5a}{y_1}\implies y_1^2-y_2y_1-5a^2=0$$

Solve for $y_1$ in terms of known $a$ and $y_2$,

$$y_1 = \frac12 y_2 + \frac12\sqrt{y_2^2+20a^2}$$

Then, the length of AD can be calculated as,

$$ED^2= AE^2 - AD^2 = (y_1-y_2)^2-a^2 = \frac14(\sqrt{y_2^2+20a^2}-y_2)^2-a^2$$

or

$$ED = \left( 4a^2 +\frac12y_2^2-\frac12 y_2 \sqrt{y_2^2+20a^2}\right)^{1/2}$$

Also, the angle $\angle ABC$ is given by

$$\sin\angle ABC = \frac{AC}{AB} = \frac{y_1}{5a} =\frac1{10}\left(\frac{y_2}{a}+ \sqrt{\left(\frac{y_2}{a}\right)^2+20} \right)$$

Answer 2

From similar triangles $ ADE,AOB $

$$ \tan A = \frac{x_1}{y_1}, \cot A =\frac{a}{DE}$$

Multiply and transpose

$$ DE= A \,x_1/y_1$$

Same way if $\frac45$ th segment is taken from top.