Question

Prove that the rhombus circumscribing a circle is a square
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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.

Hope it helps!!

Thank you