91Ó°ÊÓ

The perpendicular bisector to a line is the line that is perpendicular to the given line and also passes through the midpoint of the given line.

Theorem 4-5 states that if a point lies on the perpendicular bisector of a segment, then the point is equidistant from the endpoints of the segment.

The diagram is:

From the diagram, it can be noticed that the line $I$ is the perpendicular bisector of the line $BC$and $A$is a point lying on the perpendicular bisector of the line $BC$.

In thetriangles$∆AXB$ and $∆AXC$, it can be noticed that:

$$\begin{gathered} ∠AXB=∠AXC=90°(AX\mathrm{is}\mathrm{perpendicular}\mathrm{bisector}) \\ BX=XC(AX\mathrm{is}\mathrm{perpendicular}\mathrm{bisector}) \\ AXâ‰\dots AX\left(common\right) \end{gathered}$$

Therefore, it can be noticed that $AXâ‰\dots AX,∠AXBâ‰\dots ∠AXC$and $BXâ‰\dots XC$.

Therefore, the triangles $∆AXB$and $∆AXC$are congruent triangles by using the SAS postulate.

Therefore, by using the corresponding parts of congruent triangles it can be said that $ABâ‰\dots AC.$

That implies the point $A$is equidistant from the points $B$and $C$.

Therefore, if a point lies on the perpendicular bisector of a segment, then the point is equidistant from the endpoints of the segment.

Therefore, Theorem 4-5 is proved.