Question

prove by contradiction that point ( x ), which lies on the perpendicular bisector of ( overline{mn} ), is equidistant from endpoints ( m ) and ( n ) of the line segment.
complete the sentences by selecting the correct answers from the drop - down menus.
assume that ( overline{xm} ) and ( overline{xn} ) are choose... then show that ( \triangle xmy ) and ( \triangle xny ) are choose... by choose... this means that ( overline{xm} ) and ( overline{xn} ) choose... which contradicts the original assumption. therefore, point ( x ) on the perpendicular bisector must be equidistant from the endpoints of ( overline{xm} ).

Explanation:

Step1: Assume the opposite

Assume that \(\overline{XM}\) and \(\overline{XN}\) are not equal.

Step2: Analyze triangles

Let \(Y\) be the mid - point of \(\overline{MN}\). Since \(XY\) is the perpendicular bisector of \(\overline{MN}\), \(MY = NY\) and \(\angle XYM=\angle XYN = 90^{\circ}\), and \(XY\) is common to both \(\triangle XMY\) and \(\triangle XNY\). Then show that \(\triangle XMY\) and \(\triangle XNY\) are congruent by SAS (Side - Angle - Side).

Step3: Derive contradiction

By the congruence of \(\triangle XMY\) and \(\triangle XNY\), we have \(XM=XN\) (corresponding parts of congruent triangles are equal). So \(\overline{XM}\) and \(\overline{XN}\) are equal which contradicts the original assumption.

Answer:

Assume that \(\overline{XM}\) and \(\overline{XN}\) are not equal. Then show that \(\triangle XMY\) and \(\triangle XNY\) are congruent by SAS. This means that \(\overline{XM}\) and \(\overline{XN}\) are equal which contradicts the original assumption.