In the adjoining figure, is the center of a circle. If and are chords of the circle such that and prove that
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.