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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.

Given: In figure, a diameter AB of a circle bisects a chord PQ. AQ || PB.
To Prove: The chord PQ is also a diameter of the circle.
Proof: ∠AQP = ∠ABP    ...(1)
| Angles in the same segment
∵ AQ || PB and QP intersects them
∴ ∠AQP = ∠QPB    ...(2)
| Alt. Int. ∠s
From (1) and (2),
∠ABP = ∠QPB
⇒    ∠OBP = ∠OPB
∴ OP = OB    ...(3)
| Sides opp. to equal angles
Again,
⇒ ∠BPQ = ∠BAQ    ...(4)
| Angles in the same segment
∵ AQ || PB
and AB intersects them
∴ ∠BPQ = ∠PQA    ...(5)
| Alt. Int. ∠s
From (4) and (5),
∠BAQ = ∠PQA
⇒ ∠OAQ = ∠OQA
∴ OQ = OA    ...(6)
| Sides opp. to equal angles
Adding (3) and (6), we get
OP + OQ = OB + OA
⇒ PQ = AB
∵ AB is a diameter of the circle
∴ PQ is also a diameter of the circle.