Answer:
The two triangles are congruent, so any point on CD will be equidistant from endpoints of AB.
Step-by-step explanation:
Let the consider the figure as per the attached image:
AB be a line whose perpendicular bisector line is CD.
CD divides the line AB in two equal line segments making an angle of [tex]90^\circ[/tex] on both the sides as shown in the attached figure.
Let a point on CD be E.
Here, two triangles are formed:
[tex]\triangle AED\ and \ \triangle BED.[/tex]
Side ED is common between the two triangles.
Also, Side ED is perpendicular bisector:
[tex]\angle EDA = \angle EDB =90^\circ[/tex]
And Sides AD = DB
According to SAS congruence (i.e. Two sides are equal and angle between them is equal):
[tex]\triangle ADE \cong \triangle BDE[/tex]
And as per the properties of congruent triangles, all the sides are equal.
[tex]\Rightarrow EA = EB[/tex]
EA and EB is the distance of point E on line CD from the endpoints of line AB.
Hence proved that Any point on CD is equidistant from the endpoints of AB .
