$AB$ 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 $B$.
(N/A) Line $l$ is the perpendicular bisector of $AB$ and passes through $C$,which is the mid-point of $AB$ (see figure).
We need to show that $PA = PB$.
Consider $\Delta PCA$ and $\Delta PCB$.
In these triangles:
$AC = BC$ ($C$ is the mid-point of $AB$)
$\angle PCA = \angle PCB = 90^\circ$ (Given,as $l$ is the perpendicular bisector)
$PC = PC$ (Common side)
Therefore,by the $SAS$ congruence rule,$\Delta PCA \cong \Delta PCB$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$PA = PB$,which means $P$ is equidistant from $A$ and $B$.
We need to show that $PA = PB$.
Consider $\Delta PCA$ and $\Delta PCB$.
In these triangles:
$AC = BC$ ($C$ is the mid-point of $AB$)
$\angle PCA = \angle PCB = 90^\circ$ (Given,as $l$ is the perpendicular bisector)
$PC = PC$ (Common side)
Therefore,by the $SAS$ congruence rule,$\Delta PCA \cong \Delta PCB$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$PA = PB$,which means $P$ is equidistant from $A$ and $B$.
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