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. Show that the line PQ is the perpendicular bisector of AB.

Answer 1

PA = PB and QA = QB .....(given)
We have to show that PQ is perpendicular on AB and PQ bisects AB.
Let PQ intersect AB at C.
Let us take ΔPAQ and ΔPBQ.
In these triangles,
AP = BP , AQ = BQ ......... (Given)
PQ = PQ .......... (Common side)
So, ΔPAQ ≅ ΔPBQ ........(SSS rule)
Therefore, ∠APQ = ∠BPQ ........(CPCT).
Now let us consider ΔPAC and ΔPBC.
AP = BP ......(Given)
∠APC = ∠BPC (∠APQ = ∠BPQ already proved )
PC = PC ........(Common side)
So, ΔPAC ≅ ΔPBC ........(SAS rule)

Therefore, AC = BC (CPCT) ........... (1)
and ∠ACP = ∠BCP (CPCT)
Also, ∠ACP + ∠BCP = 180° (Linear pair)
So, 2∠ACP = 180°
or, ∠ACP = 90° ........... (2)
From (1) and (2), we can conclude that PQ is the perpendicular bisector of AB.