prove that cyclic rhombus is square
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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
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