**General enunciation**: We have to show that the middle points of equal chords of a circle are concyclic.

**Particular enunciation**: Consider, O is the centre of the circle ABCD, AB, CD and EF are three equal chords of it. M, N and P are the middle points of AB, CD and EF respectively. Let us prove that M, N and P are concyclic.

**Construction**: Let us join (O, M), (O, N) and (O, P).

**Proof**: Since M is the middle point of AB; OM is the joining line of the centre and the middle point of the chord AB, then OM ⊥ AB.

Similarly, OP⊥CD and ON ⊥ EF.

Now we know that equal chords of the circle equidistant from the circle.

∴ OM = ON = OP

Hence, if we draw a circle with the centre at O and OM or ON or as radius, it pass through the points M, N and P.

M, N and P are concyclic. (Proved)

**Problem: Show that the equal chords drawn from two ends of the diameter on its opposite sides are parallel.**

**General enunciation**: We have to prove that the equal chords drawn from two ends of the diameter on its opposite sides are parallel.

**Particular Enunciation**: Let, O is the centre of the circle ABCD and Ac is diameter. AB and CD are two equal chords are situated opposite sides of AC. We have to show that AB || CD.

**Construction**: (O, B) and (O, D) are joined.

**Proof**: In **∆ **OAB and **∆ **OCD we get,

OA = OC [radii of the same circle]

OB = OD [radii of the same circle]

and AB = CD [according to question]

∴ ∆ OAB ≅ ∆ OCD

∴ ∠ OAB = ∠OCD

i.e., ∠ BAC = ∠ACD, but these two are alternate angles and lie in the opposite sides of AC.

Hence, AC is secant of AB and CD.

∴ AB || CD

**Problem: Show that the two parallel chords of a circle drawn from two ends of a diameter on its opposite sides are equal.**

**General Enunciation:** We have to show that the two parallel chords of a circle drawn from two ends of a diameter on its opposite sides are equal.

**Particular Enunciation**: Let, O is the centre of the circle ABCD and AC is diameter. AB and CD are two parallel chords situated opposite sides of AC. Let us show that AB = CD.

**Construction**: Let us draw two perpendiculars OM and ON to the chords AB and CD respectively.

**Proof**: Since O is the centre of the circle and OM ⊥ AB, then we get, AM = ½ AB. Similarly, O is the centre of the circle and ON ⊥ CD, then we get, CN = ½ CD.

Now in right angles ∆ OAM and ∆ OCN we get,

∠OMA = ∠ ONC

∠OAM = ∠ OCN

And OA = OC

∴ ∆ OAM ≅ ∆ OCN

Then we get, AM = CN

∴ ½ AB = ½ CD

i.e., AB = CD (Showed)