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.

The theorem 4-6 states that if a point is equidistant from the endpoints of the segment then the point lies on the perpendicular bisector of a segment.

The diagram is:

From the diagram, it can be noticed that the point $A$is equidistant from the points role="math" localid="1649159940408" $B$and $C$and $X$is the midpoint of the line segment $BC$.

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

$$\begin{gathered} AB=AC(Aisequidis\tan tfromthepointsBandC) \\ BX=XC(XismidpointofBC) \\ AXâ‰\dots AX\left(common\right) \end{gathered}$$

Therefore, it can be noticed that$AX=AX,ABâ‰\dots AC$ and $BXâ‰\dots XC$.

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

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

From the diagram, it can be noticed that the angles $∠AXB$and $∠AXC$are adjacent congruent angles and the measure of adjacent congruent angles is $90°$.

Therefore, it can be noticed that $∠AXB=∠AXC=90°$

Therefore, it can be said that $AXâŠ\yen BC$.

Therefore, as $AXâŠ\yen BC$and $BXâ‰\dots XC$, therefore $AX$is perpendicular bisector of $BC$.

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

Therefore, Theorem 4-6 is proved.