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Circles

Short Answer Type

61. Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see figure). Prove that ∠ACP = ∠QCD.

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62. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lies on the third side.

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63.
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠ CAD = ∠ CBD.

Given: ABC and ADC are two right triangles with common hypotenuse AC.
To Prove: ∠CAD = ∠CBD.

Proof: ∵ AC is the common hypotenuse, ABC and ADC are two right triangles.
∴ ∠ABC = 90° = ∠ADC
⇒ Both the triangles are in the same semicircle.
∴ Points A, B, D and C are concyclic.
∵ DC is a chord
∴ ∠CAD = ∠CBD.
| ∵ Angles in the same segment are equal.

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64.

Prove that a cyclic parallelogram is a rectangle.

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Long Answer Type

65. ABCD is a cyclic trapezium with AD || BC. If ∠B = 70°, determine other three angles of the trapezium.

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66. In figure, a diameter AB of a circle bisects a chord PQ. If AQ || PB, prove that the chord PQ is also a diameter of the circle.

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Short Answer Type

67. In figure, AB = CD. Prove that BE = DE and AE = CE, where E is the point of intersection of AD and BC.

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Long Answer Type

68. ABCD is a cyclic quadrilateral with AD || BC. Prove that AB = DC.

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Short Answer Type

69. In the given figure, ABCD is a cyclic quadrilateral whose diagonals intersect at P. If ∠DBC = 70° and ∠BAC = 30°, find ∠BCD.

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70. In the given figure, ABCD is a cyclic quadrilateral whose side AB is a diameter of the circle through A, B, C, D. If ∠ADC = 140°, find ∠BAC.

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Previous Year Papers

Test Series

Test Yourself

Short Answer Type

63. ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠ CAD = ∠ CBD.

Long Answer Type

Short Answer Type

Long Answer Type

Short Answer Type

63.
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠ CAD = ∠ CBD.