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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.

Given: ABCD is a cyclic trapezium with AD || BC. ∠B = 70°.
To determine: Other three angles of the trapezium.
Determination:
∠B + ∠D = 180°
| ∵ Opposite angles of a cyclic quadrilateral are supplementary

⇒ 70° + ∠D =180°
⇒    ∠D = 180° - 70°
⇒    ∠D = 110°
Again, ∵ AD || BC and transversal AB intersects them
∵ ∠A + ∠B = 180°
| ∵    The sum of the consecutive interior angles on the same side of a transversal is 180°
⇒ ∠A + 70° = 180°
⇒    ∠A = 180° - 70°
⇒    ∠A = 110°
Also, ∠A + ∠C = 180°
| ∵    Opposite angles of a cyclic quadrilateral are supplementary
⇒ 110° + ∠C = 180°
⇒    ∠C = 180° - 110°
⇒    ∠C = 70°.

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