Show that if two chords of a circle bisect one another they must be diameters.

Explanation

Given : `AB` and `CD` are two chords of a circle, intersecting at `O` such that `OA` `=` `OB` and `OC= OD`.
To Prove : `AB` and `CD` are diameters of the circle.
Construction : Join `AC, AD` and `BC, BD`
Proof : In `Delta AOC ` and `Delta BOD`, we have
`OA` `=` `OB` [given]
`OC` `=` `OD` [given ]
`/_ AOC = /_ BOD ` [ vert. opp. `Delta `]
`:. Delta AOC ~= Delta BOD`
`implies AC= BD`
` implies hat ( AC) =hat(BD)` ....(i)

In `Delta AOD` and `Delta BOC`, we have
`OA=OB` [ given]
`OD =OC `[given ]
`/_AOD= /_BOC` [ vert. opp. `Delta`]
`:. Detla AOD ~=Delta BOC`
`implies AD =BC` ...(ii)
`implies hat (AD) = hat(BC)`.
From (i) and (ii), we get
`hat(AC)+hat(AD)= hat(BD) +hat(BC)`
`implies hat(CAD) = hat(CBD)`
`implies CD` divided the circle into two semicircles
`implies` `CD` is a diameter.
Similarly , `AB` is a diameter.