Q.Two equal chords of a circle are equidistant from the center.

Given: O is the center of the circle, and AB and CD are two equal chords. The perpendicular distances from the center O to chords AB and CD are OE and OF, i.e., \( OE \perp AB \) and \( OF \perp CD \). To prove: \( OE = OF \). Construction: Join \( O, A \) and \( O, C \). Proof: Since \( OE \perp AB \) and \( OF \perp CD \) (Given), ∴ \( AE = \cfrac{1}{2} AB \) and \( CF = \cfrac{1}{2} CD \) [Since the perpendicular from the center to any chord bisects the chord.] Again, \( AB = CD \) (Given). ∴ \( AE = CF \) ----(i) Now, in right-angled \( \triangle AEO \) and right-angled \( \triangle CFO \), \( \angle OEA = \angle OFC \) [Each is a right angle]. Hypotenuse \( OA = OC \) [Radii of the same circle]. \( AE = CF \) [From (i)]. \( \triangle AEO \cong \triangle CFO \) [By R-H-S congruence criterion]. ∴ \( OE = OF \) (Proved).
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