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Question -

ABCD is a cyclic quadrilateral whose diagonalsintersect at a point E. If ∠DBC= 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC,find ∠ECD.



Answer -

Since angles in the same segment of a circle are equal.
∴ ∠BAC = ∠BDC
⇒ ∠BDC = 30°

llso, ∠DBC= 70° [Given]
In ∆BCD, we have
∠BCD + ∠DBC + ∠CDB = 180° [Sum of anglesof a triangle is 180°]
⇒ ∠BCD + 70° + 30° = 180°
⇒ ∠BCD = 180° -100° = 80°
Now, in ∆ABC,
AB = BC [Given]
∴ ∠BCA = ∠BAC [Angles opposite toequal sides of a triangle are equal]
⇒ ∠BCA = 30° [∵ ∠B AC = 30°]
Now, ∠BCA + ∠BCD = ∠BCD
⇒ 30° + ∠ECD = 80°
⇒ ∠BCD = 80° – 30° = 50°

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