Question

$AB$ 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

Line $l\perp AB$ and passes through $C$ which is the midpoint of $AB$

To show that $PA=PB$

In $△PCA$ and $△PCB$

$AC=BC$ since $C$ is the mid-point of $AB$

$\angle PCA=\angle PCB={90}^{\circ }$(given)

$PC=PC$(common)

So,$△PCA\cong △PCB$ by SAS rule.

and so $PA=PB$ as they are corresponding sides of congruent triangles.