Question

if a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment. we let ( p ) be any point on line ( l ), but different from point ( x ). given: ( overline{px} ) is the perpendicular bisector of ( overline{wy} ) prove: ( wp = yp )

Explanation:

Step1: Definition of perpendicular bisector

Since \(PX\) is the perpendicular bisector of \(WY\), we have \(WX = YX\) (by the definition of a bisector) and \(\angle PXW=\angle PXY = 90^{\circ}\) (by the definition of perpendicular lines).

Step2: Common side

\(PX\) is a common side for \(\triangle PXW\) and \(\triangle PXY\), so \(PX=PX\) (reflexive property of equality).

Step3: SAS (Side - Angle - Side) congruence

In \(\triangle PXW\) and \(\triangle PXY\), we have \(WX = YX\), \(\angle PXW=\angle PXY\), and \(PX = PX\). By the SAS congruence criterion, \(\triangle PXW\cong\triangle PXY\).

Step4: Corresponding parts of congruent triangles

Since \(\triangle PXW\cong\triangle PXY\), then \(WP = YP\) (corresponding parts of congruent triangles are equal).

Answer:

\(WP = YP\) is proved as above.