Ina circle with centre O, AB is the diameter and CD is a chord such that ABCD is a trapezium. If $$\angle$$BAC =$$18^\circ$$, then $$\angle$$CAD is equal to:

$$\angle\ BAC\ =\ 18$$

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

ABCD is a trapezium, so AB is parallel to CD

$$\angle\ ACD=\angle\ BAC=18$$ (Alternate angle)

$$\angle\ BCD=\angle\ ACB+\angle\ ACD=90+18=108$$

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

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

$$\angle\ BAD+108=180$$

$$\angle\ BAD=72$$

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

$$72=18+\angle\ CAD$$

$$\angle\ CAD=72-18=54$$