Circles

Draw a perpendicular OM on line AD.
BC is the chord of the smaller circle and AD is the chord of the bigger circle.

We know that perpendicular drawn from the centre of the circle bisects the chord.

BM = MC ... (1)

And, AM = MD ... (2)

On subtracting equation (1) from (12, we get

AM − BM = MD − MC

AB = CD

Let OL and OM be two perpendiculars from centre O to chords AB and CD
respectively. So L and M are the mid points of chords AB and CD.

Now AB = CD (Given) ...(1)

AB/2 = CD/2 or LB = MD

In OLE and OME

OL = OM (Equal Chords are equidistant from the centre)

OLE = OME (90° each)

OE = OE (common side)

So by RHS

OLE OME

This gives, LE = ME ( by CPCT)

Here LE = LB + BE and ME = MD + DE which gives,

LB + BE = MD + DE

But LB = MD (proved above)

So, BE = DE ... (2)

Adding (1) and (2), we get

AB + BE = CD + DE

Therefore, AE = CE

Given: A circle with centre O and chord AB = chord CD.
Chords AB and CD intersect each other at E.
To Prove: AE = CE and EB = ED