Advertisements
Advertisements
प्रश्न
Prove that a cyclic parallelogram is a rectangle.
Advertisements
उत्तर

Let ABCD be a cyclic parallelogram.
∠A + ∠C = 180° (Opposite angles of a cyclic quadrilateral) ...(1)
We know that opposite angles of a parallelogram are equal.
∴ ∠A = ∠C and ∠B = ∠D
From equation (1),
∠A + ∠C = 180°
⇒ ∠A + ∠A = 180°
⇒ 2 ∠A = 180°
⇒ ∠A = 90°
Parallelogram ABCD has one of its interior angles, which is 90°. Therefore, it is a rectangle.
APPEARS IN
संबंधित प्रश्न
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.

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 = __________
ABCD is a cyclic quadrilateral in BC || AD, ∠ADC = 110° and ∠BAC = 50°. Find ∠DAC.
Prove that the circles described on the four sides of a rhombus as diameters, pass through the point of intersection of its diagonals.
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 such that ∠ADB = 30° and ∠DCA = 80°, then ∠DAB =
In the given figure, O is the centre of the circle such that ∠AOC = 130°, then ∠ABC =

If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral so formed is cyclic.
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.

