Question

AB is a line segments P and Q are points an opposite sides of AB. such that beach of them is equidistant from the points A and B. Show that the line PQ is the perpendicular bisector of AB​

Answer 1

Answer:

90

Step-by-step explanation:

Given P is equidistant from points A and B

PA=PB      

and Q is equidistant from points A and B

QA=QB        

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.