Question

Example 7: 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 (see Fig. 7.37).
Show that the line PQ is the perpendicular bisector
of AB.​

Answer 1

Answer:

Step-by-step explanation

Given P is equidistant from points A and B

PA=PB        .....(1)

and Q is equidistant from points A and B

QA=QB        .....(2)

In △PAQ and △PBQ

AP=BP  from (1)

AQ=BQ from (2)

PQ=PQ  (common)

So, △PAQ≅△PBQ  (SSS congruence)

Hence ∠APQ=∠BPQ by CPCT

In △PAC and △PBC

AP=BP from (1)

∠APC=∠BPC from (3)

PC=PC  (common)

△PAC≅△PBC  (SAS congruence)

∴AC=BC by CPCT

and ∠ACP=∠BCP by CPCT   ....(4)

Since, AB is a line segment,

∠ACP+∠BCP=180  

 (linear pair)

∠ACP+∠ACP=180  

 from (4)

2∠ACP=180  

∠ACP=  180/2

  =90  

Thus, AC=BC and ∠ACP=∠BCP=90  

 

∴,PQ is perpendicular bisector of AB.

Hence proved.