Question

prove that a cyclic rhombus is a square

Answer 1

we know that a rhombus is a parallelogram and a parallelogram has its opposite sides equal
so in a rhombus ABCD A=C and B=D
now given ABCD is a cyclic quadrilateral
and a cyclic quadrilateral has its opposite sides sum as 180
so
A+C=180
since A=C so
A+A=180
A=90Now if a rhombus has one of its angles as90 then it's a square
so ABCD is a square

Answer 2

GIVEN: Rhombus ABCD is inscribed in a circle

TO PROVE: ABCD is a SQUARE.

For proving a rhombus is a square, we just need to prove that any one of its interior angles =90° OR its diagonals are equal. Either of these….

PROOF: diagonal DB is bisector of angleB & angleD. ( As ABCD is a rhombus, So triangleABD is congruent to triangle CBD by SSS congruence criterion)

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

=> 2(a+b) = 180°

=> a+b = 90°

=> In triangle ABD

=> angleA = 180 - (a+b)

=> angleA = 180–90= 90°

So, rhombus ABCD becomes a Square…

PROVED