In the given figure , two equal chord `AB` and `CD` of a circle `C(O,r)` when produced meet at a point E.
Prove that `(i) BD = DE`
`(ii) AE = CE `

Explanation

Join `OE`.
Draw `OL` `_|_` ` AB` and `OM` `_|_` `CD`.
We know that equal chords of a circle are equidistant from the centre.
`:. OL = OM`.
In the right `Delta `OLE and OME, we have
`OL` `=` `OM`
`OE =OE` [common]
`:. Delta OLE ~=Delta OME` [ by RHS - congruence ]
`implies LE =ME` .....(i) [c.p.c.t.].
Also, `AB = CD implies (1)/(2)AB=(1)/(2)CD`
`implies BL=DM` ....(ii)
On subracting (ii) from (i), we get
`LE-BL=ME-DMimpliesBE=DE`. [Part (i) proved]
Now, AB=CD and BE`=` DE
`implies AB+BE=CD+DE`
`implies AE=C`,
Hence, `BE =DE` and `AE =CE`.