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Given AB = AC

∴ (1/2)AB = (1/2)AC

OPAB and OQAC

∴ MB = NC

⇒ ∠PMB = ∠QNC [90°]

Equal chords are equidistant from the center.

⇒ OM = ON

OP = OQ

⇒ OP – OM = OQ – ON

⇒ PM = QN

∴ ΔABC ≅ ΔABC [By side-angle-side criterion of congruence]

∴ PB = QC Proved.