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# 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)

$$\therefore$$ $$\angle{CED}$$ = $$\angle{AED}$$ = $$50^\circ$$
(Vertically opposite angles)

Also, $$\angle{ABD}$$ = $$\angle{ACD}$$
(Since, the angles in the same segment are equal)
$$\Rightarrow$$ $$\angle{ABE}$$ = $$\angle{ECD}$$
$$\therefore$$ $$\angle{ABE}$$ = $$180^\circ$$ ...(ii)

Now, in $$\triangle{CDE}$$,
$$\angle{BAC}$$ + $$20^\circ$$ + $$50^\circ$$ = $$180^\circ$$
(from eq.(i) and (ii))

$$\Rightarrow$$ $$\angle{BAC}$$ + $$70^\circ$$ = $$180^\circ$$
$$\Rightarrow$$ $$\angle{BAC}$$ = $$180^\circ$$ - $$70^\circ$$
$$\therefore$$ $$\angle{BAC}$$ = $$110^\circ$$