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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.

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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.

Given: ABCD is a cyclic quadrilateral whose side AB is a diameter of the circle through A,
B, C, D. Also,
∠ADC = 140°.

Required: To find ∠BAC.
Determination: ∵ ABCD is a cyclic quadrilateral,
∴ ∠ABC + ∠ADC = 180°
| ∵ Opposite angles of a cyclic quadrilateral are supplementary
⇒ ∠ABC + 140°= 180°
⇒    ∠ABC = 180° - 140° = 40°
Also, ∠ACB = 90°
| ∠ in a semi-circle
Now, in ∆ABC,
∠BAC + ∠ABC + ∠ACB = 180°
| ∵ The sum of the three angles of a ∆ is 180°
⇒    ∠B AC + 40° + 90° = 180°
⇒    ∠BAC + 130° = 180°
⇒ ∠B AC = 180° - 130° = 50°.

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