In the below-shown figure if AP=BP and AQ=BQ, then AM should be equal to BM.
In the given figure,
PQ is common.
So, ΔPAQ ≅ ΔPBQ (SSS rule).
Also, PA=PB (Given) and PM is common
Hence, △PMA ≅ △PMB (SAS Rule)
So, AM=BM and ∠PMA = ∠PMB (CPCT)
∠PMA + ∠PMB = 180∘ (Linear pair axiom),
∠PMA = ∠PMB = 90∘
Therefore, PM, that is, PMQ is the perpendicular bisector of AB.