Question

prove that rhombus inscribed in a circle is a square

Answer 1

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.

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 it's interior angle is 90°

Hence, rhombus inscribed in a circle is a square.

Answer 2

above is the correct answer