Question

ab is a line segment p and q are points on opposite sides of line segment ab such that each are equidistance from a and b​
show that
1] triangle AQP congruent to triangle BQP
2] triangle APC congruent to triangle BPC
3] PQ is 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