Question

Prove that acyclic rhombus is a square​

Answer 1

Answer:

Step-by-step explanation:

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.

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