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In the Given Figure, ∠Bad = 65°, ∠Abd = 70° and ∠Bdc = 45°. Find: (I) ∠Bcd (Ii) ∠Acb Hence, Show that Ac is a Diameter - ICSE Class 10 - Mathematics

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Question

In the given figure, ∠BAD = 65°, ∠ABD = 70° and ∠BDC = 45°. Find:
(i) ∠BCD (ii) ∠ACB
Hence, show that AC is a diameter

Solution

(i) In cyclic quadrilateral ABCD,
   ∠BCD = 180° - ∠BAD = 180° - 65° = 115°
   (pair of opposite angles in a cyclic quadrilateral are supplementary)

(ii) By angle sum property of ∆ABD,
    ADB = 180° - 65° - 70° = 45°
    Again, ∠ACB = ∠ADB = 45°
    (Angle in the same segment)
    ∴ ∠ADC = ∠ADB + ∠BDC = 45° + 45° = 90°
    Hence, AC is a semicircle.
    (since angle in a semicircle is a right angle)

  Is there an error in this question or solution?
Solution In the Given Figure, ∠Bad = 65°, ∠Abd = 70° and ∠Bdc = 45°. Find: (I) ∠Bcd (Ii) ∠Acb Hence, Show that Ac is a Diameter Concept: Arc and Chord Properties - Angles in the Same Segment of a Circle Are Equal (Without Proof).
S
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