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

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