Question

Given: PQ is perpendicular bisector of side AB of the triangle ABC. 

Prove: Q is equidistant from A and B.

Answer 1

Construction: Join AQ 

Proof: In ΔAQP and ΔBQP

AP = BP  ...(Given)

∠QPA = ∠QPB  ...(Each = 90°)

PQ = PQ   ...(Common)

By side – Angle – side criterion of congruence, we have

ΔAQP ≅ ΔBQP ...(SAS postulate)

The corresponding parts of the triangle are congruent

∴ AQ = BQ  ...(C.P.C.T.)

Hence Q is equidistant from A and B.