Question

In given figure $AB$ 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 $B$

Answer 1

Given AB is a line segment and l is perpendicular bisector. if point p lies on l.

Join P to A and B

The perpendicular l passe through line AB, cut on C

Then C is the midpoint of AB

Then $AC=BC$

In $\Delta PCA$ And $\Delta PCB$

$AC=BC$ (Given M is the Midpoint of AB)

$\angle PCA=\angle PCB={90}^0$ (l is the perpendicular on AB)

$PC=PC$

$\therefore \Delta PCA\cong \Delta PCB$

$\therefore PA=PB$ $\left[henceproved\right]$