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Questions
In the following figure, ∠ADC = 130° and chord BC = chord BE. Find ∠CBE.
In the given figure, AB is a diameter of the circle with centre O. If ∠ADC = 130° and chord BC = chord BE, find ∠CBE.
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Solution
Let us consider the points A, B, C, and D, which form a cyclic quadrilateral.
∴ ∠ADC + ∠OBC = 180° ... [The sum of opposite angles of a cyclic quadrilateral is 180°.]
⇒ 130° + ∠OBC = 180°
⇒ ∠OBC = 180° – 130° = 50°
In ΔBOC and ΔBOE,
BC = BE ...[Given]
OC = OE ... [Radii of a same circle]
And OB = OB ...[Common]
∴ ΔBOC ≅ ΔBOE ...[By SSS congruency]
⇒ ∠OBC = ∠OBE ...[By C.P.C.T]
Now, ∠CBE = ∠CBO + ∠EBO
= 50° + 50°
= 100°
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