**Given : **AB and CD are two equal chords of a circle which meet at E.

**To prove : **AE = CE and BE = DE

**Construction :** Draw OM ⊥ AB and ON ⊥ CD and join OE.

**Proof :** In ∆OME and ∆ONE

OM = ON [Equal chords are equidistant]

OE = OE [Common]

∠OME = ∠ONE [Each equal to 90°]

∴ ∆OME ≅ ∆ONE [RHS axiom]

⇒ EM = EN ...(i) [CPCT]

Now AB = CD [Given]

⇒ 1/ 2 AB = 1/ 2 CD

⇒ AM = CN ..(ii) [Perpendicular from centre bisects the chord]

Adding (i) and (ii), we get

EM + AM = EN + CN

⇒ AE = CE ..(iii) Now, AB = CD ..(iv)

⇒ AB – AE = CD – AE [From (iii)]

⇒ BE = CD – CE** Proved.**