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.
