# 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 .

**Proof: **

Â

Consider the and .

(radii of the same circle)

(given)

(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 âŠ¥ AB and O'F âŠ¥ CD.

**To Prove: **AB = CD

**Construction: **Join AO and CO'**Â Â Â Â
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â **

**Proof :** In right angled triangles OAE and O'CF,

OE = O'F (Given)

OA = O'C = rÂ Â Â Â Â Â Â Â Â Â (Congruent circles)

âˆ´ OAEÂ â‰… O'CFÂ Â Â Â Â Â Â Â Â (RHS congruence)

Hence, AE = CF, 2AE = 2CF

AB = CD [Perpendicular from the centre on the chord bisects the chord]