Question

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 (See Fig ). Show that the line PQ is the perpendicular bisector of AB.

Answer 1

You are given that PA = PB and QA = QB and you have to show that PQ is perpendicular on AB and PQ bisects AB.
Let PQ intersect AB at C.
$\text{Let us take}△PAQ\text{and}△PBQ.$
In these triangles,
AP = BP (Given)
AQ = BQ (Given)
PQ = PQ (Common side)
$So,△PAQ\cong △PBQ\text{(SSS rule)}$
$\text{Therefore,}\angle APQ=\angle BPQ\text{(CPCT).}$
$\text{Now let us consider}△PAC\text{and}△PBC.$
You have : AP = BP (Given)
$\angle APC=\angle BPC(\angle APQ=\angle BPQ\text{proved above)}$
PC = PC (Common side)
$\text{So,}△PAC\cong △PBC\text{(SAS rule)}$
$\text{Therefore,}AC=BC\text{(CPCT) ........... (1)}$
$\text{and}\angle ACP=\angle BCP\text{(CPCT)}$
$\text{Also,}\angle ACP+\angle BCP={180}^{\circ }\text{(Linear pair)}$
$\text{So,}2\angle ACP={180}^{\circ }$
$\text{or,}\angle ACP={90}^{\circ }\text{........... (2)}$
From (1) and (2), you can easily conclude that PQ is the perpendicular bisector of AB.