Question

AB is a line segement. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (See Fig. 12.26). Show that the line PQ is perpendicular bisector of AB.

Answer 1

Given : AB is a line segment. P and Q are points such that they are equidistant from A and B i.e. PA = PB and QA = QB

AP, PB, QA, QB, PQ are joined

To prove : PQ is perpendicular bisector of AB

Proof : In $\Delta PAQand\Delta PBQ,$

$PA=PB\left(Given\right)QA=QB\left(Given\right)PQ=PQ\left(Common\right)\therefore \Delta PAQ\cong \Delta PBQ\left(SSSaxiom\right)\therefore \angle APQ=\angle BPQ(c.p.c.t.)Nowin\Delta APC=\Delta BPCPA=PB\left(Given\right)\angle APC=\angle BPC\left(Proved\right)PC=PC\left(common\right)\therefore \Delta APC\cong \Delta BPC\left(SASaxiom\right)\therefore AC=BC(c.p.c.t.)and\angle PCA=\angle PCB(c.p.c.t.)But\angle PCA+\angle PCB={180}^{\circ }\left(Linearpair\right)\therefore \angle PCA=\angle PCB={90}^{\circ }\therefore PCorPQ\text{is perpendicular bisector of AB}$