Q.Prove that two equal chords of a circle are equidistant from the center.

Given: AB and CD are two equal chords of a circle with center O. The perpendicular distances from the center to the chords are OE and OF respectively, i.e., OE ⊥ AB and OF ⊥ CD. To prove: OE = OF Construction: Join O to A and O to C. Proof: OE ⊥ AB and OF ⊥ CD (Given) ∴ AE = ½ AB and CF = ½ CD [Because a perpendicular from the center of a circle to a chord (that is not a diameter) bisects the chord] Also, AB = CD (Given) ∴ AE = CF — (i) Now, in right-angled triangles △AEO and △CFO: - ∠OEA = ∠OFC [Each is a right angle] - Hypotenuse OA = Hypotenuse OC [Radii of the same circle] - AE = CF [From (i)] ∴ △AEO ≅ △CFO [By RHS congruence rule] ∴ OE = OF (Proved)
Similar Questions