In a circle, chords AD and BC meet at a point E outside the circle. If $$\angle$$BAE = 76$$^\circ$$ and $$\angle$$ADC= 102$$^\circ$$ , then $$\angle$$AEC is equal to:

In cyclic quadrilateral ABCD, sum of opposite angles = 180$$^\circ$$

$$=$$>  $$\angle$$BAE + $$\angle$$BCD = 180$$^\circ$$

$$=$$>   76$$^\circ$$ + $$\angle$$BCD = 180$$^\circ$$

$$=$$>   $$\angle$$BCD = 104$$^\circ$$

From the figure,

$$\angle$$ADC + $$\angle$$EDC = 180$$^\circ$$

$$=$$>  102$$^\circ$$ + $$\angle$$EDC = 180$$^\circ$$

$$=$$>  $$\angle$$EDC = 78$$^\circ$$

$$\angle$$BCD + $$\angle$$ECD = 180$$^\circ$$

$$=$$> 104$$^\circ$$ + $$\angle$$ECD = 180$$^\circ$$

$$=$$> $$\angle$$ECD = 76$$^\circ$$

In $$\triangle\ $$CDE,

$$\angle$$DEC + $$\angle$$ECD + $$\angle$$EDC = 180$$^\circ$$

$$=$$>  $$\angle$$AEC + 76$$^\circ$$ + 78$$^\circ$$ = 180$$^\circ$$

$$=$$>  $$\angle$$AEC + 154$$^\circ$$ = 180$$^\circ$$

$$=$$>  $$\angle$$AEC = 26$$^\circ$$

Hence, the correct answer is Option C