Question

$AB$ is a line segment $P$ and $Q$ are point on opposite sides of $AB$ such that each of them is equidistant from the points $A$ and $B$. Show that the line $PQ$ is perpendicular bisector of $AB$.

Answer 1

Given $P$ is equidistant from points $A$ and $B$

$PA=PB$ .....$\left(1\right)$

and $Q$ is equidistant from points $A$ and $B$

$QA=QB$ .....$\left(2\right)$

In $△PAQ$ and $△PBQ$

$AP=BP$ from $\left(1\right)$

$AQ=BQ$ from $\left(2\right)$

$PQ=PQ$ (common)

So, $△PAQ\cong △PBQ$ (SSS congruence)

Hence $\angle APQ=\angle BPQ$ by CPCT

In $△PAC$ and $△PBC$

$AP=BP$ from $\left(1\right)$

$\angle APC=\angle BPC$ from $\left(3\right)$

$PC=PC$ (common)

$△PAC\cong △PBC$ (SAS congruence)

$\therefore AC=BC$ by CPCT

and $\angle ACP=\angle BCP$ by CPCT ....$\left(4\right)$

Since, $AB$ is a line segment,

$\angle ACP+\angle BCP={180}^{\circ }$ (linear pair)

$\angle ACP+\angle ACP={180}^{\circ }$ from $\left(4\right)$

$2\angle ACP={180}^{\circ }$

$\angle ACP=\frac{{180}^{\circ }}{2}={90}^{\circ }$

Thus, $AC=BC$ and $\angle ACP=\angle BCP={90}^{\circ }$

$\therefore ,PQ$ is perpendicular bisector of $AB$.

Hence proved.