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Converse of Theorem 4

Chords of a circle which are equidistant from the centre are equal.
Proof :
Given : and are perpendiculars from the centre to the chords and .

Construction : Join and .

To Prove : Chord = Chord .



Consider the and .

(radii of the same circle)


(By RHS congruency)

The two chords are equal it they are equidistance from the cnetre.

Converse of Theorem 4 (ii)

Converse: Chords of a congruent circles which are equidistant from the corresponding centres are equal.
Given: AB and CD are the chords of the congruent circles C(O, r) and C'(O', r) respectively, such that OE = O'F, where OE AB and O'F CD.
To Prove: AB = CD
Construction: Join AO and CO'    

Proof : In right angled triangles OAE and O'CF,
OE = O'F (Given)
OA = O'C = r           (Congruent circles)
OAE  O'CF          (RHS congruence)
Hence, AE = CF, 2AE = 2CF
AB = CD [Perpendicular from the centre on the chord bisects the chord]

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