To prove rhombus inscribed in a circle is a square, we need to prove that either any one of its interior angles is equal to 90° or its diagonals are equal.

∠ABD = ∠DBC = b

∠ADB = ∠BDC = a

In the figure, diagonal BD is angular bisector of angle B and angle D.

In triangle ABD and BCD,

AD = BC (sides of rhombus are equal)

AB = CD (sides of rhombus are equal)

BD = BD (common side)

△ABD ≅ △BCD. (SSS congruency)

In the figure,

2a + 2b = 180° (as, opposite angles of a cyclic quadrilateral are always supplementary)

2(a + b) = 180°

a + b = 90°

In △ABD,

Angle A = 180°-(a + b)

= 180°-90°

= 90°

Therefore, proved that one of its interior angle is 90°

Hence, rhombus inscribed in a circle is a square.