Question

In the given figure,$\angle ADC={130}^{\circ }$ and chord $BC$ = chord $BE$. Find $\angle CBE$ (in degrees)

Answer 1

In $△BCO$ and $△BEO$

$\Rightarrow$ $BC=BE$ [ Given ]

$\Rightarrow$ $\angle BCO=\angle BEO$ [ Base angles of equal sides are also equal ]

$\Rightarrow$ $BO=BO$ [ Common side ]

$\therefore$ $△BCO\cong △BEO$ [ By $SAS$ congruence rule ]

$\Rightarrow$ $\angle CBO=\angle OBE$ [ C.P.C.T. ]

Quadrilateral $ABCD$ is a cyclic quadrilateral since, its points lies on a circle.

$\Rightarrow$ $\angle ADC+\angle CBA={180}^o$ [ Sum of opposite angles of a cyclic quadrilateral is supplementary ]

$\Rightarrow$ ${130}^o+\angle CBA={180}^o$

$\Rightarrow$ $\angle CBA={50}^o$ i.e. $\angle CBO={50}^o$

$\therefore$ $\angle OBE={50}^o$

$\Rightarrow$ $\angle CBE={50}^o+{50}^o={100}^o.$