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
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre?
In the figure m(arc LN) = 110°,
m(arc PQ) = 50° then complete the following activity to find ∠LMN.
∠ LMN = `1/2` [m(arc LN) - _______]
∴ ∠ LMN = `1/2` [_________ - 50°]
∴ ∠ LMN = `1/2` × _________
∴ ∠ LMN = __________
If the two sides of a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are equal.
In the given figure, O is the centre of the circle such that ∠AOC = 130°, then ∠ABC =
ABCD is a cyclic quadrilateral. M (arc ABC) = 230°. Find ∠ABC, ∠CDA, and ∠CBE.
In the following figure, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. Find the value of ∠ACD + ∠BED.
If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle.
The three angles of a quadrilateral are 100°, 60°, 70°. Find the fourth angle.
