# Chords equidistant from the centre of a circle are equal

## 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) Digital STUDY Center

Digital Study Center offers an effective and amazing learning platform for keen learn students in the world. We identify the needs and demands of the keen learn students which is why we stand out unique in the crowd.