Question

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

Answer 1

We have, $\angle ADC$$={130}^{\circ }$ and $BC=BE.$
Consider the points A, B, C and D form a cyclic quadrilateral.
We know that the sum of opposite angles of a cyclic quadrilateral is ${180}^{\circ }$.
$\therefore \angle ADC+\angle OBC={180}^{\circ }$
$\Rightarrow {130}^{\circ }+\angle OBC={180}^{\circ }$
$\Rightarrow \angle OBC={180}^{\circ }-{130}^{\circ }={50}^{\circ }$

In $\Delta BOC$ and $\Delta BOE$
$BC=BE$ [given equal chords]
$OC=OE$ [radii of a circle]
$OB=OB$ [common]
$\therefore \Delta BOC\cong \Delta BOE$ [by SSS congruence rule]
$\Rightarrow \angle OBC=\angle OBE={50}^{\circ }$ [by CPCT]
Now, $\angle EBO=\angle CBO={50}^{\circ }$
$\angle CBE=\angle CBO+\angle EBO$
$={50}^{\circ }+{50}^{\circ }={100}^{\circ }$