Mathematics
In the given figure AC is the diameter of the circle with center O. CD is parallel to BE.
∠AOB = 80° and ∠ACE = 20°. Calculate :
(a) ∠BEC
(b) ∠BCD
(c) ∠CED
Circles
ICSE Sp 2025
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Answer
(a) Join AE.
We know that,
The angle subtended by a chord at the centre is twice the angle subtended on the circumference.
∴ ∠AOB = 2∠AEB
⇒ 80° = 2∠AEB
⇒ ∠AEB = = 40°.
We know that,
Angle in a semi-circle is a right angle.
⇒ ∠AEC = 90°
From figure,
⇒ ∠BEC = ∠AEC - ∠AEB = 90° - 40° = 50°.
Hence, ∠BEC = 50°.
(b) From figure,
⇒ ∠ECD = ∠CEB = 50° (Alternate angles are equal)
We know that,
The angle subtended by a chord at the centre is twice the angle subtended on the circumference.
⇒ ∠AOB = 2∠BCA
⇒ 80° = 2∠BCA
⇒ ∠BCA = = 40°.
From figure,
⇒ ∠BCD = ∠BCA + ∠ACE + ∠ECD = 40° + 20° + 50° = 110°.
Hence, ∠BCD = 110°.
(c) As sum of opposite angles of cyclic quadrilateral = 180°.
⇒ ∠BED + ∠BCD = 180°
⇒ ∠BED = 180° - ∠BCD = 180° - 110° = 70°.
From figure,
⇒ ∠BED = ∠BEC + ∠CED
⇒ 70° = 50° + ∠CED
⇒ ∠CED = 70° - 50° = 20°.
Hence, ∠CED = 20°.
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