Q: 6    ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If  \small \angle DBC=70^{\circ}\small \angle BAC is \small 30^{\circ}, find \small \angle BCD. Further, if \small AB=BC, find \small \angle ECD.
 

Answers (1)

\angle BDC=\angle BAC       (angles in the same segment are equal )

\angle BDC= 30 \degree

In \triangle BDC,

       \angle BCD+\angle BDC+\angle DBC= 180 \degree

\Rightarrow \angle BCD+30 \degree+70 \degree= 180 \degree

\Rightarrow \angle BCD+100 \degree= 180 \degree

\Rightarrow \angle BCD=180 \degree- 100 \degree=80 \degree

If  AB = BC ,then 

    \angle BCA=\angle BAC

\Rightarrow \angle BCA=30 \degree

Here, \angle ECD+\angle BCE=\angle BCD

   \Rightarrow \angle ECD+30 \degree=80 \degree

\Rightarrow \angle ECD=80 \degree-30 \degree=50 \degree

 

 

 

 

 

 

 

 

 

 

 

 

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