Is there a purely geometrical way to prove the fact below? Given two lines intersecting and making an angle theta wrt to each other . It not hard to to show by coordinate geometry that locus of points whose sum of perpendicular distances from these lines is a constant is a rectangle but does there exists a purely geometrical way to show the same locus?
 A: Define a value $d$ corresponding to the sum of the distances from the lines $r$ and $s$. Start from a point $A$ on one line, for example $r$, such that the distance from $A$ to line $s$  is equal to $d$:

Now trace the line $s'$, parallel to $s$ and passing through $A$. Among the resulting four angles with vertex in $A$, choose one of the two angles in the semiplane containing $s$ (for example, choose the larger one). Trace the  bisector of this angle until you cross  $s$ in $B$. The points of this bisector are equidistant from $r$ and $s'$: therefore, compared to $A$, any of these points is farther from $r$ by some distance, and closer to $s$ by the same distance. Then, the sum of the distances from $r$ and $s$ is $d$ for all the points of $AB$, which is the first side of our rectangle:

Now we can repeat the same procedure by drawing the line $r'$ parallel to $r$ and passing through $B$. Again, we can consider the four angles with vertex in $B$, and look at the two ones in the semiplane containing $r$. This time we choose the smaller of these two angles (note that we have already traced the bisector of the larger one) and draw its bisector until we cross $r$. By the same considerations as above, compared to $B$, any point of this bisector is farther from $s$ by some distance, and closer to $r$ by the same distance. Then, the sum of the distances from $r$ and $s$ is $d$ for all the points of this bisector. So we get the second side of our rectangle:

Note that these two sides are perpendicular to each other, because they are bisectors of adjacent, supplementary angles.
Continuing in this way, we can draw the third and fourth side of our rectangle.
