In a circle with centre O, AB is the diameter and CD is a chord such that ABCD is a trapezium. If $$\angle$$BAC = $$40^\circ$$, then $$\angle$$CAD is equal to:

$$\angle\ BAC\ =\ 40$$

$$\angle\ ACB=90\ $$ (Angle on semicircle)

ABCD is a trapezium, so AB is parallel to CD

$$\angle\ ACD=\angle\ BAC=40$$ (Alternate angle)

$$\angle\ BCD=\angle\ ACB+\angle\ ACD=90+40=130$$

ABCD is a cyclic quadrilateral because all point lies on same circle

$$\angle\ BAD+\angle\ BCD=180$$

$$\angle\ BAD+130=180$$

$$\angle\ BAD=50$$

$$\angle\ BAD=\angle\ BAC+\angle\ CAD$$

$$50=40+\angle\ CAD$$

$$\angle\ CAD=50-40=10$$