In the given figure, AB is a line segment. P and Q are points on opposite sides 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.

Explanation

GIVEN A line segment AB and two point P and Q such that PA = PB and QA = QB.
TO PROVE Let AB and PQ intersect at C. Then, we have to prove that AC = BC and `angleACP=90^(@)`.
PROOF In `DeltaPAQ and DeltaPBQ,` we have
`PA=PB" (given)"`
`OA=QB" (given)"`
`PQ=PQ" (common)"`
`therefore" "DeltaPAQ~=DeltaPBQ" (by SSS-criteria)."`
`therefore" "angleAPQ = angleBPQ" ...(i) (c.p.c.t.)."`
Now, in `DeltaPAC and DeltaPBC,` we have
`PA=PB" (given)"`
`angleAPC=angleBPC" "[because angleAPQ=angleBPQ" in (i)"]`
`PC=PC" (common)"`
`therefore" "DeltaPAC~=DeltaPBC" (by SAS-criteria)."`
`therefore" "AC=BC" ...(ii) (c.p.c.t.)."`
And, `angleACP=angleBCP" ...(iii) (c.p.c.t.)."`
But, `angleACP+angleBCP=180^(@)" (linear pair)."`
`therefore" "2angleACP=180^(@)" [using (iii)]."`
So, `angleACP=90^(@).`
Hence, PQ is the perpendicular bisector of AB.