Que 6 plz solve this
Q.6. In the given figure, O is the centre of the circle passing through the  points A, B, C and D and DC is produced to point E. If $\angle$BAD = 60$°$, find $\angle$BCE and $\angle$BOD.

Dear Student,

We will need to make use of two theorems to solve this question.

Theorem 1 : The angle subtended by an arc at the centre is double the angle subtended by it any point on the remaning part of the circle.


Theorem 2: The sum of opposite angles in an inscribed quadrilateral in a circle is 180^o.

Now, 
It is given that ​$$\begin{gathered} \angle BAD = 60 \\ So, from Theorem 2, \\ \angle BAD + \angle BCD = 180 \\ Now, \sin ce ED is a straight line, \\ \angle BCD + \angle BCE = 180 \\ \angle BCD = 180 - \angle BCE \\ So, \\ \angle BAD + 180 - \angle BCE = 180 \\ \angle BAD = \angle BCE +180 -180 \\ \angle BAD = \angle BCE = 60° \end{gathered}$$


Also, from Theorem 1, 

$\angle BOD = 2\angle BAD = 2 \times 60 = 120°$

Thus, angle BOD is 120^o and angle BCE is 60^o.

Hope it helps.

Regards