Advertisements
Advertisements
प्रश्न
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
Advertisements
उत्तर
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.
संबंधित प्रश्न
In the given figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at 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 = __________
In a cyclic quadrilateral ABCD, if ∠A − ∠C = 60°, prove that the smaller of two is 60°
ABCD is a cyclic quadrilateral in ∠BCD = 100° and ∠ABD = 70° find ∠ADB.
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 in which BA and CD when produced meet in E and EA = ED. Prove that AD || BC .
ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that EB = EC.
In the given figure, ABCD is a quadrilateral inscribed in a circle with centre O. CD is produced to E such that ∠AED = 95° and ∠OBA = 30°. Find ∠OAC.
