5. In figure, A, B and C are four points on a circle. AC and BD intersect at a point E such that $$\angle{BEC}$$ = $$130^\circ$$ and $$\angle{ECD}$$ = $$20^\circ$$. Find $$\angle{BAC}$$.

In $$\triangle{ABC}$$,
$$\angle{AED}$$ = $$180^\circ$$ -$$130^\circ$$ = $$50^\circ$$ ...(linear pair of angles)
So, $$\angle{CED}$$ = $$\angle{AED}$$ = $$50^\circ$$ ...(Vertically opposite angles)
Also, $$\angle{ABD}$$ = $$\angle{ACD}$$ ...(Since, the angles in the same segment are equal)
i.e., $$\angle{ABE}$$ = $$\angle{ECD}$$
Thus, $$\angle{ABE}$$ = $$180^\circ$$ ...(ii)
Now, in $$\triangle{CDE}$$,
$$\angle{BAC}$$ + $$20^\circ$$ + $$50^\circ$$ = $$180^\circ$$ ...(from eq.(i) and (ii))
i.e., $$\angle{BAC}$$ + $$70^\circ$$ = $$180^\circ$$
i.e., $$\angle{BAC}$$ = $$180^\circ$$ - $$70^\circ$$
Therefore, $$\angle{BAC}$$ = $$110^\circ$$