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Question

AB is a line segment P and Q are point on opposite sides of AB such that each of them is equidistant from the points A and B. Show that the line PQ is perpendicular bisector of AB.
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Solution

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, PAQPBQ (SSS congruence)
Hence APQ=BPQ by CPCT
In PAC and PBC
AP=BP from (1)
APC=BPC from (3)
PC=PC (common)
PACPBC (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)
2ACP=180
ACP=1802=90
Thus, AC=BC and ACP=BCP=90
,PQ is perpendicular bisector of AB.
Hence proved.

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