Circle

In the figure, $AC$ is the diameter of a circle with center $O$.

$CD$ and $BE$ are parallel.

$\angle AOB = 80^\circ$ and $\angle ACE = 10^\circ$. Find:

1. $\angle BEC$;
2. $\angle BCD$;
3. $\angle CED$.

Given:

$AC =$ diameter of circle

$O =$ center of circle

$CD \parallel BE$

$\angle AOB = 80^\circ$; $\;$ $\angle ACE = 10^\circ$

1. $\angle AOB + \angle BOC = 180^\circ$ $\;\;\;$ [linear pair]
   
   $\therefore \;$ $\angle BOC = 180^\circ - \angle AOB = 180^\circ - 80^\circ = 100^\circ$
   
   Now, $\;$ $\angle BEC = \dfrac{1}{2} \angle BOC$ $\;\;\;$ [angle at center is twice the angle at the remaining circumference]
   
   i.e. $\;$ $\angle BEC = \dfrac{1}{2} \times 100^\circ = 50^\circ$



2. $CD \parallel BE$ and $EC$ is a transversal.
   
   $\therefore \;$ $\angle ECD = 50^\circ$
   
   Now, $\;$ $OB = OC$ $\;\;\;$ [radii of the circle]
   
   $\therefore \;$ $\angle OBC = \angle OCB$ $\;\;\;$ [angles opposite equal sides are equal]
   
   Now, $\;$ in $\triangle BOC$,
   
   $\angle OBC + \angle OCB + \angle BOC = 180^\circ$ $\;\;\;$ [sum of angles of a triangle]
   
   i.e. $\;$ $2 \angle OCB + \angle BOC = 180^\circ$
   
   i.e. $\;$ $2 \angle OCB = 180^\circ - \angle BOC = 180^\circ - 100^\circ = 80^\circ$
   
   $\implies$ $\angle OCB = 40^\circ$
   
   Now, $\;$ $\angle BCD = \angle ECD + \angle ACE + \angle OCB$
   
   i.e. $\angle BCD = 50^\circ + 10^\circ + 40^\circ = 100^\circ$



3. $BCDE$ is a cyclic quadrilateral.
   
   Then, $\;$ $\angle BED + \angle BCD = 180^\circ$ $\;\;\;$ [opposite angles of a cyclic quadrilateral are supplementary]
   
   $\therefore \;$ $\angle BED = 180^\circ - \angle BCD = 180^\circ - 100^\circ = 80^\circ$
   
   Now, $\;$ $\angle BED = \angle BEC + \angle CED$
   
   $\therefore \;$ $\angle CED = \angle BED - \angle BEC = 80^\circ - 50^\circ = 30^\circ$