Question -
Answer -
Given : AB is a line segment.
P and Q are points such that they are equidistant from A and B
i.e. PA = PB and QA = QB AP, PB, QA, QB, PQ are joined
To prove : PQ is perpendicular bisector of AB
Proof : In ∆PAQ and ∆PBQ,
PA = PB (Given)
QA = QB (Given)
PQ = PQ (Common)
∴ ∆PAQ ≅ ∆PBQ (SSS axiom)
∴ ∠APQ = ∠BPQ (c.p.c.t.)
Now in ∆APC = ∆BPC
PA = PB (Given)
∆APC ≅ ∆BPC (Proved)
PC = PC (Common)
∴ ∆APC = ∆BPC (SAS axiom)
∴ AC = BC (c.p.c.t.)
and ∠PCA = ∠PCB (c.p.c.t.)
But ∠PCA + ∠PCB = 180° (Linear pair)
∴ ∠PCA = ∠PCB = 90°
∴ PC or PQ is perpendicular bisector of AB