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# 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,

$$\angle{BDC}$$ + $$\angle{DBC}$$ + $$\angle{BCD}$$ = $$180^\circ$$
$$\Rightarrow$$ $$30^\circ$$ + $$70^\circ$$ + $$\angle{BCD}$$ = $$180^\circ$$
$$\Rightarrow$$ $$100^\circ$$ + $$\angle{BCD}$$ = $$180^\circ$$
$$\Rightarrow$$ $$\angle{BCD}$$ = $$180^\circ$$ - $$100^\circ$$
$$\therefore$$ $$\angle{BCD}$$ = $$80^\circ$$

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 $$\angle{BCA}$$ = $$30^\circ$$)

$$\Rightarrow$$ $$\angle{ECD}$$ = $$30^\circ$$
$$\therefore$$ $$\angle{EDC}$$ = $$30^\circ$$