Question 139687
I think (and this is just my humble opinion, mind you) that you are making this way too difficult.


Given, line segment AB and line segment PQ that is the perpendicular bisector of AB.  Let Point X be the point of intersection of the two line segments.


Choose any point R on the segment PQ.  Construct segments RA and RB.

Then RX = RX by identity, AX = BX by the definition of a perpendicular bisector.  Angle RXA is a right angle and Angle RXB is a right angle by definition of perpendicular.  Two triangles with two equal sides and equal included angles are congruent by some theorem or another -- you look it up.  Therefore RA = RB, but RA is the distance from R to A and RB is the distance from R to B.  Therefore R is equidistant from A and B.  QED.


{{{drawing(600,600,0,10,0,10,grid(1),
line(1,5,9,5),
line(5,8,5,1),
line(1,5,5,7),
line(9,5,5,7),
red(locate(1,5,A),
locate(9,5,B),
locate(5,8,P),
locate(5,1,Q),
locate(5,5,X)),
green(locate(5,7,R))

)}}}