Advertisements
Advertisements
Question
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
Advertisements
Solution
Let ABCD be a rhombus in which diagonals are intersecting at point O and a circle is drawn while taking side CD as its diameter. We know that a diameter subtends 90° on the arc.
∴ ∠COD = 90°
Also, in rhombus, the diagonals intersect each other at 90°.
∠AOB = ∠BOC = ∠COD = ∠DOA = 90°
Clearly, point O has to lie on the circle.
RELATED QUESTIONS
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
In the given figure, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠DBC = 55° and ∠BAC = 45°, find ∠BCD.
Prove that the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.
ABCD is a cyclic quadrilateral. M (arc ABC) = 230°. Find ∠ABC, ∠CDA, and ∠CBE.
Find all the angles of the given cyclic quadrilateral ABCD in the figure.
In a cyclic quadrilaterals ABCD, ∠A = 4x, ∠C = 2x the value of x is
ABCD is a cyclic quadrilateral such that ∠A = 90°, ∠B = 70°, ∠C = 95° and ∠D = 105°.
If non-parallel sides of a trapezium are equal, prove that it is cyclic.
