Advertisements
Advertisements
प्रश्न
ABCD is a cyclic quadrilateral in which AB is parallel to DC and AB is a diameter of the circle. Given ∠BED = 65°, calculate:
- ∠DAB,
- ∠BDC.

ABCD is a cyclic quadrilateral and DC || AB. If AB is the diameter of the circle and BED = 65°, find:
- ∠DAB
- ∠BDC

Advertisements
उत्तर
(i) Find ∠DAB
Angles BED and DAB are angles subtended by the same chord DB in the same circle.
∴ ∠DAB = ∠BED = 65°
(ii) Find ∠BDC
Since AB is a diameter, angle ADB is an angle in a semicircle:
∠ADB = 90°.
In triangle ABD:
∠ABD = 180° − (∠ADB + ∠DAB)
∠ABD = 180° − (90° + 65°)
∠ABD = 25°.
AB ∥ DC,
∠ABD and ∠BDC are corresponding angles.
∴∠BDC = ∠ABD = 25°.
संबंधित प्रश्न
In the figure, m∠DBC = 58°. BD is the diameter of the circle. Calculate:
1) m∠BDC
2) m∠BEC
3) m∠BAC

Prove that the rhombus, inscribed in a circle, is a square.
Two circles intersect at P and Q. Through P diameters PA and PB of the two circles are drawn. Show that the points A, Q and B are collinear.
In the figure, given alongside, AB || CD and O is the centre of the circle. If ∠ADC = 25°; find the angle AEB. Give reasons in support of your answer.

In the given figure, PQ is a diameter. Chord SR is parallel to PQ. Given that ∠PQR = 58°,
Calculate:
- ∠RPQ,
- ∠STP.

In the following figure, AD is the diameter of the circle with centre O. Chords AB, BC and CD are equal. If ∠DEF = 110°, calculate: ∠AEF

Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°.
Calculate : ∠DBC
Also, show that the ΔAOD is an equilateral triangle.

In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°.
Calculate: ∠ADC
Also, show that the ΔAOD is an equilateral triangle.

In the given figure, BAD = 65°, ABD = 70°, BDC = 45°.
(i) Prove that AC is a diameter of the circle.
(ii) Find ACB.

