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°
Angle A = 180°-(a + b)
Therefore, proved that one of its interior angle is 90°
Hence, rhombus inscribed in a circle is a square.
Rate this question :
In the figure PR and QS are two diameters. Is PQ = RS?
Given that the vertices A, B, C of a quadrilateral ABCD lie on a circle. Also ∠A + ∠C = 180°, then prove that the vertex D also lie on the same circle.AP- Mathematics
A is the centre of the circle and ABCD is a square. If BD = 4cm then find the radius of the circle.
Prove that a cyclic rhombus is a square.AP- Mathematics
For each of the following, draw a circle and inscribe the figure given. If a polygon of the given type can’t be inscribed, write not possible.
(c) Obtuse triangle
(d) Non-rectangular parallelogram
(e) Accute issosceles triangle
(f) A quadrilateral PQRS with as diameter.AP- Mathematics