Where to place a bridge over the highway? I've got a problem to solve. I had 2 different ideas which didn't actually work for all cases.
On the both sides of highway there are 2 houses K and L (as in the attached picture). Line, that passes through K and L cannot be perpendicular to the highway. Where should we build a bridge (perpendicular to the highway borders), that straigh tracks from both houses to the highway would be the same (a and a in the picture)? Use compasses and a ruler to find a place to build the brige.
I attached a sample schema to illustrate the problem.

I'd be glad if you gave me some clues or complete solution, since I'm extremely curious how to solve it.
Best regards,
Tom.
 A: Let us say that the highway runs due east-west.
Construct a new point $L'$ that is "due north" of point $L$ and the distance between $L$ and $L'$ is the width of the highway. In effect, we have removed the highway from the plane and moved the half-plane below the lower edge of the highway up to the upper edge of the highway.
The point you want on the line (that used to be the highway) is equidistant from $L'$ and $K$. Just draw the perpendicular bisector of line segment $\overline{KL'}$ and mark its intersection with the upper edge of the highway. Then drop a perpendicular segment from there to the lower edge of the highway, draw segments from the highway points to $K$ and $L$, and you are done.

In the construction above, parallel lines $a$ and $b$ (the highway) and points $K$ and $L$ (the houses) are given (all in blue).
Drop line $c$ through $L$ and perpendicular to $a$. Mark points $A$ and $B$ on lines $a$ and $B$. Use a compass to draw circle $d$ centered at $L$ with radius $AB$. Mark point $L'$ on line $c$ and circle $d$ between $A$ and $L$.
Draw line $e$ the perpendicular bisector of segment $\overline{KL'}$. Mark point $C$ the intersection of $b$ and $e$. (This intersection is guaranteed to exist, due to the restriction that the line $\overline{KL}$ cannot be perpendicular to the highway.) Draw line $f$ through $C$ perpendicular to $b$, and mark point $D$ the intersection of $f$ and $a$.
Your desired path is then $K$ to $C$ to $D$ to $L$ (all in red).
