Question

In given figure $AB$ is a line-segment. $P$ and $Q$ are points on either side of $AB$ such that each of them is equidistant from the points $A$ and $B$. Show that the line $PQ$ is the perpendicular bisector of $AB$

Answer 1

Given AB is a line segment .P and Q are two-point on either side of AB such That $AP=BPandAQ=BQ$

In $\Delta PAQ$and $\Delta PBQ$

$AP=BP$ (Given )

$AQ=BQ$ (Given )

$PQ=PQ$ (Common side)

$\therefore \Delta PAQ\cong \Delta PBQ$

$\therefore \angle APQ=\angle BPQ$

In $\Delta PAC$ and $PBC$

$AP=BP$ (Given )

$\angle APC=\angle BPC$ We proved above $\angle APQ=\angle BPQ$

$PC=PC$ (Common Side)

$\therefore \Delta PAC\cong PBC$

$\therefore AC=BC$....................................(1)

$\therefore \angle ACP=\angle BCP$

$\angle ACP+\angle BCP={180}^0$

$\therefore 2\angle ACP={180}^0$

$\Rightarrow \angle ACP={90}^0$.....................(2)

From (1) and (2) the line PQ is the perpendicular bisector of AB. $\left[henceproved\right]$