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+1 vote
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in Circles by (23.7k points)
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ACB is a tangent to a circle at C . CD and CE are chords such that ∠ACE > ∠ACD. If ∠ACD = ∠BCE = 50°, then

(a) CD = CE 

(b) ED is not parallel to AB 

(c) ED passes through the center of the circle 

(d) ∆ CDE is right angled triangle.

1 Answer

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by (22.8k points)
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Best answer

Answer : CD = CE 

∠ACD = ∠CED = 50° (Alternate Segment Theorem) 

∠BCE = ∠CDE = 50° 

⇒ ∠D = ∠E = 50° 

CD = CE 

Also, ∠ACD = ∠D = 50°, but these are alternate interior angles 

⇒ ED || AB 

∠DCE =180° – (50° + 50°) =  80° 

∴  ΔCDE is an acute angled triangle.

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