Proving in a triangle $AB$ and $CD$ are two straight lines  intersecting in $O$. $XY$ is another straight line.  Show that in general two points can be found on $XY$ which are equidistant from $AB$ and $CD$. But isn't there only one such point?
Please help. I know we can find plenty of such points on the anglular bisector of $AB$ and $CD$, but how about when the line $XY$ doesn't coincide  with the angular bisector?
 A: The lines $AB$ and $CD$ have two angular bisectors. They are the lines $l$ and $m$ in the sketch below. So the two points you're looking for are the intersections of $XY$ with the two angular bisectors.  
In general  any line wich 


*

*doesn't coincide with $AB$ or $CD$

*isn't perpendicular to either $l$ or $m$

*doesn't go through $O$,


will have exactly two points wich are equidistant to $AB$ and $CD$.
N.B. You really should talk about the line through $A$ and $B$ or rename it $l$ (or some other lowercase letter) otherwise they are line segments as I mentioned.

Sorry for the low quality sketch, but I'm not on my home computer, 
so I was having a hard time with it. This seemed like the best solution to me.
A: AB and CD will have an angular bisectors PQ and RS.  Every point on PQ and RS will be equidistant from AB and CD.  
An arbitary line XY will intersect PQ only once; unless XY is either colinear (infinitely many intersections) or parallel (no intersections) to PQ.  Such a point of intersection will be fit the required criteria.
Likewise with RS.  XY cannot be parallel to both RS and PQ unless AB and CD are colinear.
There is one more case.  XY can intercept both PQ and RS through O.
In any other case, there will be exactly two points on XY that fit the required criteria.
