Question

AB is a line segment and line l is its perpendicular bisector. Show that every point on line l is equidistant from A and B.

Answer 1

In the following figure, AB is the given line segment and l is its perpendicular bisector.
A any point on the given line l.


Now, in ∆AOP and∆BOP
$$\begin{gathered} AO=BO \\ \angle AOP=\angle BOP={90}^{\circ }\left(l\mathrm{is}\mathrm{perpendicular}\mathrm{bisector}\mathrm{of}AB\right) \\ OP=OP\left(\mathrm{Common}\right) \end{gathered}$$
$$\begin{gathered} \therefore ∆AOP\cong ∆BOP \\ \Rightarrow AP=BP\left(\mathrm{By}\mathrm{c}.\mathrm{p}.\mathrm{c}.\mathrm{t}\right) \end{gathered}$$
Hence, every point on line l is equidistant from A and B.