Question

please give me answer​

Answer 1

in triangle BCO and B0E
CO = OE (radii of same circle)
bo = bo ( common)
BC = BE(given)
so by SSS congruence rule 
triangle BCO is congruent to triangle BOE
BY CPCT
angle CBO = angle OBE ( eqn 1)
since abcd is a cyclic quad.
therefore angle D + angle B = 180 
130 + angle B = 180
angle B = 50 degree
angle CBO + angle OBE = angle CBE
BUt angle cbo = angle OBE
therefore 50 + 50 = angle CBE
angle CBE = 100 degree

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Answer 2

Hi Mate,

The point A, B, C and D formed a cyclic quadrilateral .

∴ ∠ADC + ∠OBC = 180°

⇒ 130°+ ∠OBC = 180°

⇒ ∠OBC = 180°

Now, in ΔBOC and ΔBOE ,

BC = BE [given]

OC = OE [radii of the same circle]

OB = OB [common side]

∴ ΔBOC ≅ ΔBOE [by SSS congruent rule]

Then , ∠OBC = ∠OBE = 50° [CPCT]

∴ ∠CBE = ∠CBO + ∠EBO = 50° + 50° = 100°

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