 6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If $$\angle{DBC}$$ = $$70^\circ$$, $$\angle{BAC}$$ = $$30^\circ$$, Find $$\angle{BCD}$$. Further, if AB = BC, find $$\angle{EDC}$$. As, Angles in the same segment are equal, we get,
$$\angle{BDC}$$ = $$\angle{BAC}$$
Therefore, $$\angle{BDC}$$ = $$30^\circ$$.
In $$\triangle{DBC}$$, we have,
$$\angle{BDC}$$ + $$\angle{DBC}$$ + $$\angle{BCD}$$ = $$180^\circ$$
$$30^\circ$$ + $$70^\circ$$ + $$\angle{BCD}$$ = $$180^\circ$$
i.e., $$100^\circ$$ + $$\angle{BCD}$$ = $$180^\circ$$
i.e., $$\angle{BCD}$$ = $$180^\circ$$ - $$100^\circ$$
Therefore, $$\angle{BCD}$$ = $$80^\circ$$

If AB= BC, $$\angle{BCA}$$ = $$\angle{BAC}$$ = $$30^\circ$$ ...(Since, Angles opposite to equal sides in a traingle are equal)
Now, $$\angle{ECD}$$ = $$\angle{BCD}$$ - $$\angle{BCA}$$ = $$80^\circ$$ - $$30^\circ$$ ...(Given $$\angle{BCD}$$ = $$80^\circ$$ and $$\angle{BCA}$$ = $$30^\circ$$)
i.e., $$\angle{ECD}$$ = $$30^\circ$$
Therefore, $$\angle{EDC}$$ = $$30^\circ$$