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 (CPCT)
Hence Q is equidistant from A and B.