`A C\ a n d\ B D` are chords of a circle which bisect each other. Prove that (i) `A C\ a n d\ B D` are diameters (ii) `A B C D` is a rectangle

Explanation

We conclude from the given information that `AB` and `CD` are the two chords of a circle
Construction
Let the point of intersection be `O`.
Join AB,BC,CD and AD.

In triangles `AOB` and `COD`
,
`/_AOB=/_COD ` ...(Vertically opposite angles)

`OB=OD` ....(`O` is the mid-point of `BD`)
`OA=OC` ....(`O` is the mid-point of `AC`)
`/_\AOB~=/_\COD` ....`SAS` test of congruence
`:.AB=CD` .... (by c.s.c.t.)
Similarly, we can prove
`/_\AOD~=/_\BOC,`
then we get

`:.AD=BC` ....(by c.s.c.t.)
So, `squareABCD` is a parallelogram, since opposite sides are equal in length.
So, opposite angles are equal as well.
So, `/_A=/_C`
Also, for a cyclic quadrilateral opposite angles add up to `180^@`So,
`/_A+/_C=180^@`
`/_A+/_A=180^@`
`/_A=90^@`
So, `BD` is the diameter. Similarly, `AC` is also the diameter.
Since `AC` and `BD` are diameters,
`:./_A=/_B=/_C=/_D=90^@` ...(Angle inscribed in a semi circle is a right angle)
Hence, parallelogram `ABCD` is a rectangle.