In figure, \(\angle{ABC}\) = \(69^\circ\), \(\angle{ACB}\) = \(31^\circ\). Find \(\angle{BDC}\).

As we know that, the angles in the same segment are equal.
We get,

\(\angle{BDC}\) = \(\angle{BAC}\) ...(i)

Now, in \(\triangle{ABC}\),

\(\angle{BAC}\) + \(\angle{ABC}\) + \(\angle{ACB}\) = \(180^\circ\)
\(\Rightarrow \) \(\angle{BAC}\) + \(69^\circ\) + \(31^\circ\) = \(180^\circ\)
\(\Rightarrow \)\(\angle{ABC}\) + \(100^\circ\) = \(180^\circ\)
\(\Rightarrow \) \(\angle{ABC}\) = \(180^\circ\) - \(100^\circ\)
\(\therefore \) \(\angle{BAC}\) = \(80^\circ\)
\(\therefore \) \(\angle{BDC}\) = \(80^\circ\) ...(from (i))