Question

AB is Aline segment and l is its perpendicular bisector.show AP=BP.P is a point on l

Answer 1

Answer: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=  

2

180  

∘

 

​  

=90  

∘

 

Thus, AC=BC and ∠ACP=∠BCP=90  

∘

 

∴,PQ is perpendicular bisector of AB.

Hence proved.

Step-by-step explanation:

Answer 2

Answer:

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