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प्रश्न
In the given figure, C and D are points on the semi-circle described on AB as diameter. Given angle BAD = 70° and angle DBC = 30°, calculate angle BDC.
उत्तर
Since ABCD is a cyclic quadrilateral, therefore, ∠BCD + ∠BAD = 180° ...(Since opposite angles of a cyclic quadrilateral are supplementary)
`=>` ∠BCD + 70° = 180°
`=>` ∠BCD = 180° − 70° = 110°
In ΔBCD, we have,
∠CBD + ∠BCD + ∠BDC = 180°
`=>` 30° + 110° + ∠BDC = 180°
`=>` ∠BDC = 180° − 140°
`=>` ∠BDC = 40°
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संबंधित प्रश्न
In the figure, ABCD is a cyclic quadrilateral with BC = CD. TC is tangent to the circle at point C and DC is produced to point G. If ∠BCG = 108° and O is the centre of the circle, find :
- angle BCT
- angle DOC
In the given figure, AB is the diameter of a circle with centre O. ∠BCD = 130°. Find:
(i) ∠DAB
(ii) ∠DBA
D and E are points on equal sides AB and AC of an isosceles triangle ABC such that AD = AE. Prove that the points B, C, E and D are concyclic.
In a square ABCD, its diagonals AC and BD intersect each other at point O. The bisector of angle DAO meets BD at point M and the bisector of angle ABD meets AC at N and AM at L. Show that:
- ∠ONL + ∠OML = 180°
- ∠BAM + ∠BMA
- ALOB is a cyclic quadrilateral.
In the figure, O is the centre of the circle and the length of arc AB is twice the length of arc BC. If angle AOB = 108°, find: ∠ADB.
In the given below the figure, O is the centre of the circle and ∠ AOC = 160°. Prove that 3∠y - 2∠x = 140°.
ABCD is a cyclic quadrilateral AB and DC are produced to meet in E. Prove that Δ EBC ∼ Δ EDA.
Prove that the angle bisectors of the angles formed by producing opposite sides of a cyclic quadrilateral (Provided they are not parallel) intersect at the right angle.
Prove that the quadrilateral formed by angle bisectors of a cyclic quadrilateral ABCD is also cyclic.
An exterior angle of a cyclic quadrilateral is congruent to the angle opposite to its adjacent interior angle, to prove the theorem complete the activity.
Given: ABCD is cyclic,
`square` is the exterior angle of ABCD
To prove: ∠DCE ≅ ∠BAD
Proof: `square` + ∠BCD = `square` .....[Angles in linear pair] (I)
ABCD is a cyclic.
`square` + ∠BAD = `square` ......[Theorem of cyclic quadrilateral] (II)
By (I) and (II)
∠DCE + ∠BCD = `square` + ∠BAD
∠DCE ≅ ∠BAD
