same distance from a point to 2 non-parallel lines

Question

There are 2 nonparallel lines $a,b$ and point $E$ which doesn't belong to any of them and lies anywhere between them. EDIT: Task is to find two couples of points F, G and H, I $\in$ y such that $|EF|=|FG|$ and $|EH|=|HI|$. (where $|FG|$ and $|HI|$ are minimal distances between lines $a,b$ with given conditions) There are two solutions. I don't know how to construct points F, G and H, I with compass and ruler.

Thanks for your advice.

EDIT: Solution looks like this on the picture: more properties and details on the same construction:

I need help with construction steps of points F, G and H, I. I don't understand how computer found it.

Answer 1

Like this?

given by linesPt[-3, 11, .4 x + 2.5, -.2 x + 1.5], code here. ie basically, by solving for when $GF$ and $IH$ meet certain requirements.

Answer 2

This will only work if the skew lines form an acute angle. Obtuse will obviously not work.

What you have described by, "$|EF|=|FG|$ and $|EH|=|HI|$**" are the two equal length sides of isosceles triangels. That means that the base is parallel to the line being touched by the top of the isosceles triangle.

So: this should work for the isoceles closer to the intersection.

step one: draw a line parallel to the line you wish to touch with the top of your isoceles triangle, that goes through the point of interest "E" as you named it.

step two: identify the intersection of this line and the other skew line: the segment between the point E and this intersection is the base of the isoceles.

step three: draw a perpendicular bisector of this "base".

step four: identify the intersection of this line and the other parllel line, which will form the top point of the isoceles.

step five: draw the sides, which should be obvious now.

Now you have your first triangle.

Not sure about the second.

Answer 3

Use the directrix definition of a parabola, i.e. the parabola is the locus of all points that have the same distance to a given point as a given line.

So, the problem reduces to finding the two (potential) intersection points of the line $a$ with the parabola with focus $E$ and directrix $b$. See here for explicit constructions of the intersection points.