# 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 $\small \angle ECD$.

S seema garhwal

$\angle BDC=\angle BAC$       (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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