Advertisements
Advertisements
प्रश्न
Prove that the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.
Advertisements
उत्तर
Here, ABCD is a cyclic rectangle; we have to prove that the centre of the corresponding circle is the intersection of its diagonals.
Let O be the centre of the circle.
We know that the angle formed in the semicircle is 90°.
Since, ABCD is a rectangle, So
`angleADC = angleDCB = angleABC = angleBAD = 90°`
Therefore, AC and BD are diameter of the circle.
We also know that the intersection of any two diameter is the centre of the circle.
Hence, the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.
APPEARS IN
संबंधित प्रश्न
In the given figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
Prove that a cyclic parallelogram is a rectangle.
AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters; (ii) ABCD is a rectangle.
Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90°-A, 90° − `1/2 A, 90° − 1/2 B, 90° − 1/2` C.
In the figure, `square`ABCD is a cyclic quadrilateral. Seg AB is a diameter. If ∠ ADC = 120˚, complete the following activity to find measure of ∠ BAC.
`square` ABCD is a cyclic quadrilateral.
∴ ∠ ADC + ∠ ABC = 180°
∴ 120˚ + ∠ ABC = 180°
∴ ∠ ABC = ______
But ∠ ACB = ______ .......(angle in semicircle)
In Δ ABC,
∠ BAC + ∠ ACB + ∠ ABC = 180°
∴ ∠BAC + ______ = 180°
∴ ∠ BAC = ______
ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that EB = EC.
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.
If a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are also equal.
If non-parallel sides of a trapezium are equal, prove that it is cyclic.
