Chords equidistant from the centre of a circle are equal
Proposition:Let AB and CD be two chords of a circle with centre O. OE and OF are the perpendiculars from O to the chords AB and CD respectively. Then OE and OF represent the distances from centre to the chords AB and CD respectively.If OE = OF, it is to be proved that AB = CD.
Construction:
Join O,A and O,C.Step-1: Since OE⊥AB and OF⊥CD (right angles)
Therefore, ∠OEA = ∠OFC = 1 right angle.
Step-2: Now, between the right-angled
ΔOAE and ΔOCF
hypotenuse OA = hypotenuse OC (radius of same circle)
and OE = OF. (supposition)
∴ ΔOAE ≅ ΔOCF
∴ AE = CF.
Step-3: AE = 1/2 AB and CF = 1/2 CD. (Perpendicular from the centre)
Step-4: Therefore, 1/2 AB = 1/2 CD
i.e., AB = CD. (Proved)
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