Question
✓
Answer
In the given figure, we have $\angle\text{ADC}=130^\circ$ and chord BC = BE. We have to find $\angle\text{CBE}.$ Since ABCD is a cyclic quadrilateral and the opposite angles of a cyclic quadrilateral are supplementary.
$\therefore\angle\text{D}+\angle\text{ABC}=180^\circ$
$\Rightarrow130^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-130^\circ=50^\circ$
$\Rightarrow\angle\text{OBC}=50^\circ....(1)$
In $\triangle\text{OBC}$ and $\triangle\text{OBE},$ we have
BC = BE [Given]
OC = OE [Radii of same circle]
OB = OB [Common side]
$\therefore\triangle\text{OBC}\cong\triangle\text{OBE}$ [By SSS cong. Rule]
$\angle\text{OBC}+\angle\text{OBE}=50^\circ$ $\big[$By C.P.C.T. and by (1) $\angle\text{OBC}=50^\circ\big]$
$\therefore\angle\text{OBC}+\angle\text{OBE}=50^\circ+50^\circ=100^\circ$
Hence, $\angle\text{CBE}=100^\circ$
$\therefore\angle\text{D}+\angle\text{ABC}=180^\circ$
$\Rightarrow130^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-130^\circ=50^\circ$
$\Rightarrow\angle\text{OBC}=50^\circ....(1)$
In $\triangle\text{OBC}$ and $\triangle\text{OBE},$ we have
BC = BE [Given]
OC = OE [Radii of same circle]
OB = OB [Common side]
$\therefore\triangle\text{OBC}\cong\triangle\text{OBE}$ [By SSS cong. Rule]
$\angle\text{OBC}+\angle\text{OBE}=50^\circ$ $\big[$By C.P.C.T. and by (1) $\angle\text{OBC}=50^\circ\big]$
$\therefore\angle\text{OBC}+\angle\text{OBE}=50^\circ+50^\circ=100^\circ$
Hence, $\angle\text{CBE}=100^\circ$
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