Statement-I      The sum of opposite angles of a cyclic quadrilateral is $180°$ .

Statement-II      If a rhombus is cyclic, then it is a square.

Statement (I) is true, by concept.

Statement (II) is true

$ABCD$  is a rhombus

$\overline{)A}=\overline{)C}$  (opposite angles of rhombus are equal)

$\because$  $ABCD$  is cyclic, $\overline{)A}+\overline{)C}=180°$

$\overline{)A}+\overline{)A}=180°$      $\left(\because \overline{)A}=\overline{)C}\right)$

$2\overline{)A}=180°$

$\overline{)A}=\frac{180°}{2}=90°$

$\therefore$  A rhombus is in which one angle is a right angle, then it is a square

$\therefore$  $ABCD$  is square.