Let ABCD is a cyclic quadrilateral in which\[AB\text{ }||\text{ }DC\] \[AC\text{ }and\text{ }BD\]are its diagonals.

Required to prove:

\[\left( i \right)\text{ }AD\text{ }=\text{ }BC\]

\[\left( ii \right)\text{ }AC\text{ }=\text{ }BD\]

Proof:

(i) As \[AB\text{ }||\text{ }DC\](given)

\[\angle DCA\text{ }=\angle CAB\] [Alternate angles]

Now, \[chord\text{ }AD\]subtends \[\angle DCA\text{ }and\text{ }chord\text{ }BC\] subtends \[\angle CAB\] at the circumference of the circle.

So,

\[\angle DCA\text{ }=\angle CAB\]

Hence, \[chord\text{ }AD\text{ }=\text{ }chord\text{ }BC\text{ }or\text{ }AD\text{ }=\text{ }BC.\]

(ii) Now, in \[\vartriangle ABC\text{ }and\text{ }\vartriangle ADB\]

\[AB\text{ }=\text{ }AB\][Common]

\[\angle ACB\text{ }=\angle ADB\] [Angles in the same segment are equal]

\[BC\text{ }=\text{ }AD\][Proved above]

Hence, by SAS criterion of congruence

\[\vartriangle ACB\cong \vartriangle ADB\]

By CPCT

\[AC\text{ }=\text{ }BD\]