In the adjoining figure, DE is a chord parallel to diameter AC of the circle with centre O

Question:

In the adjoining figure, DE is a chord parallel to diameter AC of the circle with centre O. If CBD = 60°, calculate ∠CDE.

 

Solution:

Angles in the same segment of a circle are equal.
i.e., CAD = CBD = 60°
We know that an angle in a semicircle is a right angle.
i.e., ADC = 90°
In  ΔADCwe have:
ACD + ADC + CAD = 180°  (Angle sum property of a triangle)
⇒ ACD + 90° + 60° = 180°
ACD = 180° –  (90° + 60°) = (180° – 150°) = 30°
CDE = ACD = 30°  (Alternate angles as AC parallel to DE)
Hence, CDE = 30° 

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