here's the solution,
let <u>the segment be AB</u>, and PQ is perpendicular bisector of AB at Q , and Let P be the point on perpendicular bisector, and join P with A and B,
and since <u>PQ bisects AB into AQ</u> and BQ, they are equal,
that is , AQ = BQ
now, two triangles are formed.
In the given triangles, <u>PQA and PQB</u>
( common side )
( perpendicular forms 90° )
( they are equal, proved above )
here, Triangle PQA and PQB are congruent by SAS ( side angle side ) criterion.
So, AP = BP ( by C.P.C.T )