In the given figure, equal chords `AB` and `CD` of a circle `C(O,r)` cut at right angles at `E.` If `M` and `N` are the midpoints of `AB` and `CD` respectively, prove that `OMEN` is a square.

Explanation

Given : `A` circle `C(O,r)` in which chord `AB` `=` chord `CD`, and `N` are the midpoints of `AB` and `CD` respectively.
To Prove : `OMEN` is a square.
Construction : Join `OE`
Proof : Since the line joining the centre of a circle to the midpoint of a chord is perpendicular to the chord, we have
`OM _|_ AG` and ` ON _|_ CD`
`implies OMB= 90^(@)` and `/_ OND =90^(@)`
`implies /_OME =90^(@)` and `/_ ONE =90^(@)`
Also, equal chords of a circle are equidistant from the centre.
`:. OM=ON ` ....(i)
Now, in `Delta OME` and `Delta ONE`, we have
`OM` `=` `ON` [ from (i) ]
`/_ OME = /_ ONE ` [ each equal to `90^(@)`]
`OE = OE` [ common]
`:. Delta OME ~= Delta ONE `[ by `SAS` -congruence ]
`implies ME = NE `[ c.p.c.t. ]
Thus, in qud. `OMEN`, we have
`OM=ON,ME=NE` and `/_OME =Delta ONE = 90^(@)`
Hence, `OMEN` is a square.