`A Ca n dB D` are chords of a circle that bisect each other. Prove that: `A Ca n dB D` are diameters `ABCD` is a rectangle .

Explanation

We conclude from the given information
`AB` and `CD` are the two chords of a circle
Lets assume the point of intersection be `O`.
Construction
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` ....`c.s.c.t`.
Similarly,
we can prove
`/_\AOD~=/_\BOC`, then we get
`AD=BC` ....`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.
(ii) 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..