\(AB\) is a line segment and line \(l\) is its perpendicular bisector. If a point \(P\) lies on \(l\), show that \(P\) is equidistant from \(A\) and \(B\).
Proof:
Consider \(∆ PCA\) and \(∆ PCB\).
\(AC = BC\) (\(C\) is the mid-point of \(AB\))
\(∠ PCA = ∠ PCB = \)\(^°\) (Given)
\(PC = PC\) (Common)
So, by rule, \(∆ PCA ≅ ∆ PCB\)
And so, \(PA = PB\) [By CPCT]
Hence, we proved.