Question

in given figure angleADC =130dgree and BC=BE then find angleADC CBE

Answer 1

the fig is of what shape

Answer 2

Answer:

Step-by-step explanation:

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

donot forget to make it brainlist answer.