Question

prove that a cyclic 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.

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

hi mate your answer

the opposite angles of cyclic quadrilateral are suplementry

Let the pair of opposite angles be x and y

x+y=180----(1)

the opposite angles of rhombus are congruent

x=y-----(2)

from (1)and (2)

x+x=180

2x=180

x=180/2

x=90°

the all sides and angles of cylic rhombus are congruent and the measure of each angle is 90°

the quadrilateral having all sides are congruent and each angle is 90°

hence,cyclic rhombus is a square

hope that it will help you